🧠:Okay, so I need to solve this equation: the floor of (5 + 6x)/8 equals (15x - 7)/5. Let me make sure I understand the problem first. The floor function, denoted by [A], means the greatest integer less than or equal to A. So, for example, [3.2] would be 3, and [5.9] would be 5. Similarly, if the number is already an integer, like [4], it stays 4. So, the left side of the equation is the floor of (5 + 6x)/8, and the right side is (15x - 7)/5. I need to find all real numbers x that satisfy this equation.First, let me write down the equation again to be clear:floor[(5 + 6x)/8] = (15x - 7)/5Hmm. Since the left side is a floor function, it must be an integer. Therefore, the right side, (15x - 7)/5, must also be an integer. Let's denote that integer as n. So, n is an integer such that:(15x - 7)/5 = nandfloor[(5 + 6x)/8] = nTherefore, I can set up two equations here. First, from the right-hand side:(15x - 7)/5 = n => 15x - 7 = 5n => 15x = 5n + 7 => x = (5n + 7)/15So, x must be equal to (5n + 7)/15 for some integer n. Then, plugging this into the left-hand side floor function, we need:floor[(5 + 6x)/8] = nBut since x is expressed in terms of n, let's substitute x:floor[(5 + 6*(5n + 7)/15)/8] = nSimplify the numerator inside the floor function:First, compute 6*(5n + 7)/15:6*(5n + 7) = 30n + 42, then divide by 15:(30n + 42)/15 = 2n + 14/5So, adding 5 to that:5 + 2n + 14/5 = 5 + 14/5 + 2nConvert 5 to 25/5:25/5 + 14/5 = 39/5, so the numerator becomes 39/5 + 2nTherefore, the expression inside the floor function is:(39/5 + 2n)/8 = (39 + 10n)/40So, floor[(39 + 10n)/40] = nTherefore, the equation reduces to:floor[(39 + 10n)/40] = nNow, this is interesting. So, the floor of (39 + 10n)/40 must equal n. Let's write this as:floor[(10n + 39)/40] = nBut (10n + 39)/40 is equal to (10n)/40 + 39/40 = n/4 + 39/40. So, this simplifies to:floor(n/4 + 39/40) = nSo, the left-hand side is the floor of n/4 + 39/40, which must equal n. Let's analyze this.Let me denote the argument of the floor function as y(n) = n/4 + 39/40. Then, floor(y(n)) = n.So, for each integer n, we need y(n) to be in the interval [n, n + 1), because the floor of y(n) is n. Therefore:n ≤ y(n) < n + 1Substituting y(n):n ≤ n/4 + 39/40 < n + 1Let me solve the inequalities for n.First inequality:n ≤ (n/4) + 39/40Multiply both sides by 40 to eliminate denominators:40n ≤ 10n + 3940n - 10n ≤ 3930n ≤ 39n ≤ 39/30n ≤ 13/10n ≤ 1.3Since n is an integer, this inequality holds for n ≤ 1.Second inequality:(n/4) + 39/40 < n + 1Multiply both sides by 40:10n + 39 < 40n + 4010n + 39 - 40n - 40 < 0-30n - 1 < 0-30n < 1Multiply both sides by -1 (remember to reverse the inequality):30n > -1n > -1/30Since n is an integer, this inequality holds for n ≥ 0.Therefore, combining both inequalities:0 ≤ n ≤ 1Thus, n can be 0 or 1.So, possible integer values for n are 0 and 1.Now, let's check these possible n values.Case 1: n = 0Check if floor(y(0)) = 0, where y(0) = 0/4 + 39/40 = 39/40 = 0.975floor(0.975) = 0, which is correct. So, n = 0 is valid.Case 2: n = 1Compute y(1) = 1/4 + 39/40 = 10/40 + 39/40 = 49/40 = 1.225floor(1.225) = 1, which is correct. So, n = 1 is also valid.Case 3: n = 2Wait, but according to our previous analysis, n must be 0 or 1. Let's check if n = 2 could be a solution, but according to inequalities, n should not be greater than 1.3, so n = 2 is invalid. But let's test it just in case:y(2) = 2/4 + 39/40 = 0.5 + 0.975 = 1.475. floor(1.475) = 1 ≠ 2. So, invalid.Similarly, n = -1: y(-1) = (-1)/4 + 39/40 = (-10/40) + 39/40 = 29/40 = 0.725. floor(0.725) = 0 ≠ -1. So invalid.Thus, only n = 0 and n = 1 are possible.Therefore, the solutions for x are:For n = 0:x = (5*0 + 7)/15 = 7/15 ≈ 0.4667For n = 1:x = (5*1 + 7)/15 = 12/15 = 4/5 = 0.8But wait, we need to verify these solutions in the original equation, because sometimes when dealing with floor functions, substitutions can introduce extraneous solutions. Let me check each x value.First, check x = 7/15.Compute left-hand side: floor[(5 + 6*(7/15))/8]Calculate 6*(7/15) = 42/15 = 14/5 = 2.85 + 14/5 = 5 + 2.8 = 7.8Divide by 8: 7.8 /8 = 0.975floor(0.975) = 0Right-hand side: (15*(7/15) -7)/5 = (7 -7)/5 = 0/5 = 0So, 0 = 0. Correct.Next, check x = 4/5.Left-hand side: floor[(5 + 6*(4/5))/8]Calculate 6*(4/5) = 24/5 = 4.85 + 4.8 = 9.8Divide by 8: 9.8/8 = 1.225floor(1.225) = 1Right-hand side: (15*(4/5) -7)/5 = (12 -7)/5 = 5/5 = 11 = 1. Correct.Therefore, both solutions are valid.But wait, are there any other possible x values that could satisfy the equation outside of these n values?Wait, in our approach, we considered that n must be integer because the right-hand side (15x -7)/5 is equal to n. But is there a possibility that (15x -7)/5 is an integer even if x isn't of the form (5n +7)/15? Wait, actually, solving (15x -7)/5 = n gives x = (5n +7)/15, so every solution must be of this form for some integer n. Therefore, since we found all integers n (0 and 1) that satisfy the equation floor[(5 +6x)/8] = n, then x must be 7/15 or 4/5.Therefore, the solutions are x = 7/15 and x = 4/5.But let me just make sure that there are no other possible n values that we might have missed due to miscalculations in inequalities.Let me re-examine the inequalities:Starting with n ≤ (n/4) + 39/40Multiply both sides by 40:40n ≤ 10n + 3930n ≤ 39n ≤ 39/30 = 1.3, so n ≤ 1.3. Since n is integer, n ≤ 1.Then the other inequality:(n/4) + 39/40 < n +1Multiply by 40:10n + 39 < 40n + 40-30n < 1n > -1/30 ≈ -0.0333So n ≥ 0 as integer.Hence, n can only be 0 or 1, which we checked. Therefore, these are the only solutions.Therefore, the solutions are x = 7/15 and x = 4/5.But let me check if there's another approach to solving the original equation without substituting n. Maybe directly working with inequalities.Original equation:floor[(5 +6x)/8] = (15x -7)/5Let me denote k = floor[(5 +6x)/8], which is equal to (15x -7)/5. So, k is an integer, and (15x -7)/5 = k => x = (5k +7)/15.Therefore, substituting back into the floor function:floor[(5 +6*(5k +7)/15)/8] = kWhich is the same as before:floor[(39 +10k)/40] = kSo, this leads us back to the same equation. Therefore, our previous steps are correct.Alternatively, another way is to set up inequalities for the floor function.Given that floor[(5 +6x)/8] = (15x -7)/5 = k (integer).So, the floor of (5 +6x)/8 is k, which means:k ≤ (5 +6x)/8 < k +1Multiply all parts by 8:8k ≤ 5 +6x <8k +8Subtract 5:8k -5 ≤6x <8k +8 -5Simplify:8k -5 ≤6x <8k +3Divide by 6:(8k -5)/6 ≤x <(8k +3)/6But we also have that (15x -7)/5 =k => x=(5k +7)/15Therefore, substituting x=(5k +7)/15 into the inequalities:(8k -5)/6 ≤(5k +7)/15 <(8k +3)/6Let me solve these inequalities for k.First inequality:(8k -5)/6 ≤ (5k +7)/15Multiply both sides by 30 to eliminate denominators:5*(8k -5) ≤2*(5k +7)40k -25 ≤10k +1440k -10k ≤14 +2530k ≤39k ≤39/30 =13/10=1.3Second inequality:(5k +7)/15 <(8k +3)/6Multiply both sides by 30:2*(5k +7) <5*(8k +3)10k +14 <40k +1514 -15 <40k -10k-1 <30kk >-1/30 ≈-0.0333...Since k is integer, k ≥0Therefore, combining both inequalities, k must be 0 or 1, as before.Therefore, the same conclusion.Therefore, x=(5*0 +7)/15=7/15 and x=(5*1 +7)/15=12/15=4/5.Thus, the solutions are x=7/15 and x=4/5.To make absolutely sure, let me check another way. Suppose x=7/15≈0.4667Compute left-hand side:(5 +6x)/8=(5 +6*(7/15))/8=(5 +42/15)/8=(5 +2.8)/8=7.8/8=0.975floor(0.975)=0Right-hand side:(15x -7)/5=(15*(7/15)-7)/5=(7 -7)/5=0/5=0. Correct.x=4/5=0.8Left-hand side:(5 +6*(4/5))/8=(5 +24/5)/8=(5 +4.8)/8=9.8/8=1.225floor(1.225)=1Right-hand side:(15*(4/5)-7)/5=(12 -7)/5=5/5=1. Correct.So, both check out.Therefore, the solutions are x=7/15 and x=4/5.I think that's thorough. I considered substitution, inequalities, checked both solutions, verified the possible integer values, and made sure there are no extraneous solutions. So, I'm confident these are the correct solutions.Final AnswerThe solutions are boxed{dfrac{7}{15}} and boxed{dfrac{4}{5}}.

✅:To solve the equation ( leftlfloor frac{5 + 6x}{8} rightrfloor = frac{15x - 7}{5} ), we start by noting that the right-hand side must be an integer. Let ( n ) be this integer. Thus, we have:[frac{15x - 7}{5} = n implies 15x - 7 = 5n implies x = frac{5n + 7}{15}]Substituting ( x = frac{5n + 7}{15} ) into the left-hand side of the original equation, we get:[leftlfloor frac{5 + 6 left( frac{5n + 7}{15} right)}{8} rightrfloor = n]Simplifying the expression inside the floor function:[frac{5 + 6 left( frac{5n + 7}{15} right)}{8} = frac{5 + frac{30n + 42}{15}}{8} = frac{5 + 2n + frac{14}{5}}{8} = frac{frac{39}{5} + 2n}{8} = frac{39 + 10n}{40}]Thus, the equation becomes:[leftlfloor frac{39 + 10n}{40} rightrfloor = n]We need ( frac{39 + 10n}{40} ) to be in the interval ([n, n + 1)). This leads to the inequalities:[n leq frac{39 + 10n}{40} < n + 1]Solving these inequalities:1. ( n leq frac{39 + 10n}{40} ) [ 40n leq 39 + 10n implies 30n leq 39 implies n leq frac{13}{10} approx 1.3 ]2. ( frac{39 + 10n}{40} < n + 1 ) [ 39 + 10n < 40n + 40 implies -30n < 1 implies n > -frac{1}{30} approx -0.033 ]Since ( n ) must be an integer, the possible values are ( n = 0 ) and ( n = 1 ).For ( n = 0 ):[x = frac{5(0) + 7}{15} = frac{7}{15}]For ( n = 1 ):[x = frac{5(1) + 7}{15} = frac{12}{15} = frac{4}{5}]Verification:- For ( x = frac{7}{15} ): [ leftlfloor frac{5 + 6 left( frac{7}{15} right)}{8} rightrfloor = leftlfloor 0.975 rightrfloor = 0 quad text{and} quad frac{15 left( frac{7}{15} right) - 7}{5} = 0 ]- For ( x = frac{4}{5} ): [ leftlfloor frac{5 + 6 left( frac{4}{5} right)}{8} rightrfloor = leftlfloor 1.225 rightrfloor = 1 quad text{and} quad frac{15 left( frac{4}{5} right) - 7}{5} = 1 ]Both solutions are valid. Therefore, the solutions are (boxed{dfrac{7}{15}}) and (boxed{dfrac{4}{5}}).

🔑:Given the equation (leftlfloor frac{5 + 6x}{8} rightrfloor = frac{15x - 7}{5}), where ( leftlfloor A rightrfloor ) denotes the greatest integer less than or equal to (A).1. Express the floor function in terms of an inequality: By definition of the floor function: [ leftlfloor frac{5 + 6x}{8} rightrfloor leq frac{5 + 6x}{8} < leftlfloor frac{5 + 6x}{8} rightrfloor + 1 ]2. Assign a variable to the floor term and set up equality: Let ( n = leftlfloor frac{5 + 6x}{8} rightrfloor ). Then we have: [ n = frac{15x - 7}{5} ]3. Solve for (x) in terms of (n): Rearranging the expression for (x), we get: [ n = frac{15x - 7}{5} implies 5n = 15x - 7 implies 15x = 5n + 7 implies x = frac{5n + 7}{15} ]4. Substitute (x) back into the floor inequality: Using (x = frac{5n + 7}{15}), we need to ensure: [ n leq frac{5 + 6left(frac{5n + 7}{15}right)}{8} < n + 1 ] 5. Simplify the inequality: First, let’s simplify: [ n leq frac{5 + frac{30n + 42}{15}}{8} < n + 1 ] [ n leq frac{5 + 2n + 2.8}{8} < n + 1 ] [ n leq frac{7 + 2n}{8} < n + 1 ]6. Verify both sides of the inequality: Let's break the inequality into two parts and verify: - For the left inequality ( n leq frac{7 + 2n}{8} ): [ n cdot 8 leq 7 + 2n implies 8n leq 7 + 2n implies 6n leq 7 implies n leq frac{7}{6} ] - For the right inequality (frac{7 + 2n}{8} < n + 1): [ frac{7 + 2n}{8} < n + 1 implies 7 + 2n < 8n + 8 implies 7 < 6n + 8 implies -1 < 6n implies n > -frac{1}{6} ] Considering (n) is an integer, ( n leq 1 ).7. Integer solutions for (n): The possible integer values for ( n ) that satisfy the inequalities ( -frac{1}{6} < n leq frac{7}{6} implies -frac{1}{6} < n le 1 ) are ( n = 0, 1 ). 8. Evaluate corresponding (x) values: For ( n = 0 ): [ x = frac{5 cdot 0 + 7}{15} = frac{7}{15} ] For ( n = 1 ): [ x = frac{5 cdot 1 + 7}{15} = frac{12}{15} = frac{4}{5} ]Therefore the solutions for ( x ) are ( frac{7}{15} ) and ( frac{4}{5} ). Conclusion:[boxed{x = frac{7}{15} text{ or } x = frac{4}{5}}]

🧠:Okay, so I need to evaluate the integral of sin cubed z times cos z from π/4 to π/2. Hmm, let me think. Integrals involving products of sine and cosine often use substitution. Since the power of sine is odd here, maybe substitution is a good approach. Let me recall: when you have an odd power of sine or cosine, you can often use substitution by letting u be the other function. Wait, here the integral is sin^3 z * cos z dz. So maybe let u = sin z? Because the derivative of sin z is cos z dz, which is present here. Let me check that.If I set u = sin z, then du/dz = cos z, so du = cos z dz. That means the integral becomes u^3 * du, right? Because sin^3 z becomes u^3 and cos z dz becomes du. So integrating u^3 du is straightforward. The integral of u^3 is (u^4)/4 + C. Then substitute back to get (sin^4 z)/4 evaluated from π/4 to π/2. That seems right. Let me verify.Let me write down the steps more formally. Let u = sin z. Then du = cos z dz. So when z = π/2, u = sin(π/2) = 1. When z = π/4, u = sin(π/4) = √2/2. Therefore, the integral becomes the integral from u = √2/2 to u = 1 of u^3 du. Integrating that gives [ (u^4)/4 ] from √2/2 to 1. Calculating the definite integral: (1^4)/4 - ( (√2/2)^4 )/4. Simplify that. 1/4 - ( ( (√2)^4 ) / (2^4) ) /4. Wait, (√2)^4 is (2)^(1/2 * 4) = 2^2 = 4. And 2^4 is 16. So (√2/2)^4 is 4/16 = 1/4. Therefore, the second term is (1/4)/4 = 1/16. So the integral is 1/4 - 1/16 = (4/16 - 1/16) = 3/16. So the answer should be 3/16. Let me check that again.Wait, hold on, (√2/2)^4: √2 is 2^(1/2), so (2^(1/2)/2)^4 = (2^(1/2))^4 / 2^4 = 2^(2) / 16 = 4/16 = 1/4. So yes, (√2/2)^4 is 1/4. Then divide by 4: 1/4 divided by 4 is 1/16. So 1/4 minus 1/16 is indeed 3/16. So the result is 3/16. That seems correct. Alternatively, maybe I can check by using another method. Let's see. If I didn't use substitution, but tried to integrate by parts. Wait, but integrating by parts might complicate things here. Let's see. Let me set u = sin^3 z and dv = cos z dz. Then du would be 3 sin^2 z cos z dz, and v = sin z. Then integration by parts gives uv - integral of v du. So sin^3 z * sin z - integral of sin z * 3 sin^2 z cos z dz. Wait, that becomes sin^4 z - 3 integral sin^3 z cos z dz. Wait, but that brings back the original integral. Let me call the original integral I. So, I = sin^4 z - 3I. Then, moving 3I to the left: I + 3I = sin^4 z, so 4I = sin^4 z, so I = (sin^4 z)/4 + C. Which is the same result as substitution. Therefore, the same answer. So that confirms it.Therefore, substituting the limits from π/4 to π/2 gives [ (sin^4(π/2))/4 - (sin^4(π/4))/4 ] = [ (1)/4 - ( (√2/2)^4 )/4 ] = [1/4 - (1/4)/4] = 1/4 - 1/16 = 3/16. So the integral is 3/16. Wait, but hold on, in the integration by parts approach, when evaluating the definite integral, we need to apply the limits to uv and to the integral part. But since we ended up with the same antiderivative, maybe it's consistent. Either way, both methods lead to the same result. So I think the answer is correct.Alternatively, another way to check is to compute the indefinite integral first. Let me compute ∫ sin^3 z cos z dz. Let u = sin z, du = cos z dz. So ∫ u^3 du = u^4/4 + C = sin^4 z /4 + C. Then, evaluating from π/4 to π/2 gives [sin^4(π/2)/4 - sin^4(π/4)/4] = [1/4 - ( (sqrt(2)/2)^4 )/4] = 1/4 - ( (4/16)/4 )? Wait, no, (sqrt(2)/2)^4 is ( (sqrt(2))^4 )/(2^4 ) = (4)/(16) = 1/4. So then, sin^4(π/4)/4 = (1/4)/4 = 1/16. Therefore, 1/4 - 1/16 = 3/16. Correct.Therefore, the answer is 3/16. I think that's solid. Both substitution and integration by parts confirm it. Also, the limits were computed correctly. So I feel confident that the integral evaluates to 3/16.

✅:The integral (int_{pi / 4}^{pi / 2} sin^{3} z cos z , dz) can be evaluated using substitution. Here's the step-by-step solution:1. Substitution: Let ( u = sin z ). Then, ( du = cos z , dz ).2. Adjust Limits: - When ( z = pi/2 ), ( u = sin(pi/2) = 1 ). - When ( z = pi/4 ), ( u = sin(pi/4) = sqrt{2}/2 ).3. Rewrite Integral: The integral becomes: [ int_{sqrt{2}/2}^{1} u^3 , du ]4. Integrate: [ int u^3 , du = frac{u^4}{4} + C ]5. Evaluate Definite Integral: [ left[ frac{u^4}{4} right]_{sqrt{2}/2}^{1} = frac{1^4}{4} - frac{(sqrt{2}/2)^4}{4} = frac{1}{4} - frac{1/4}{4} = frac{1}{4} - frac{1}{16} = frac{3}{16} ]Answer: (boxed{dfrac{3}{16}})

🔑:1. Identify the Integral and Strategy: We need to evaluate the integral: [ int_{frac{pi}{4}}^{frac{pi}{2}} sin^3 z cos z , dz ] Notice that the integrand is sin^3 z cos z. A useful strategy here is to use substitution.2. Substitution: Let: [ u = sin z ] Then, taking the derivative of both sides, we have: [ du = cos z , dz ] This directly helps us transform cos z , dz into du, and the integral becomes: [ int sin^3 z cos z , dz = int u^3 , du ]3. Change of Limits: We need to change the bounds of integration according to our substitution: - When ( z = frac{pi}{4} ): [ u = sinleft(frac{pi}{4}right) = frac{sqrt{2}}{2} ] - When ( z = frac{pi}{2} ): [ u = sinleft(frac{pi}{2}right) = 1 ] So, the integral in terms of ( u ) and adjusting the limits becomes: [ int_{frac{pi}{4}}^{frac{pi}{2}} sin^3 z cos z , dz = int_{frac{sqrt{2}}{2}}^{1} u^3 , du ]4. Evaluate the Integral: The integral now is: [ int_{frac{sqrt{2}}{2}}^{1} u^3 , du ] To integrate ( u^3 ), use the power rule for integration: [ int u^3 , du = frac{u^4}{4} + C ] So, evaluate the definite integral: [ left. frac{u^4}{4} right|_{frac{sqrt{2}}{2}}^{1} ]5. Calculate the Bounds: Evaluate the expression at the upper and lower limits: [ left. frac{u^4}{4} right|_{frac{sqrt{2}}{2}}^{1} = frac{1^4}{4} - frac{left(frac{sqrt{2}}{2}right)^4}{4} ] Simplify each term: [ frac{1}{4} - frac{left(frac{sqrt{2}}{2}right)^4}{4} = frac{1}{4} - frac{left(frac{2}{4}right)^2}{4} = frac{1}{4} - frac{left(frac{1}{2}right)^2}{4} = frac{1}{4} - frac{1}{16} ] Find a common denominator to subtract the fractions: [ frac{1}{4} - frac{1}{16} = frac{4}{16} - frac{1}{16} = frac{3}{16} ] Conclusion:The value of the integral is:[boxed{frac{3}{16}}]

🧠:Okay, let's try to tackle this problem step by step. First, I need to understand what's given and what needs to be found. The problem involves three pumps working together for irrigation. Here's what I know:1. The first and third pumps finished working at the same time.2. The second pump stopped working 2 hours after starting.3. The first pump pumped 9 m³ of water.4. The second and third pumps together pumped 28 m³.5. The third pump pumps 3 m³ more per hour than the first pump.6. All three pumps together pump 14 m³ per hour.We need to find the pumping rate per hour for each pump. Let's denote the rates as variables:- Let ( x ) be the rate of the first pump in m³ per hour.- Then, the third pump's rate is ( x + 3 ) m³ per hour.- Let ( y ) be the rate of the second pump in m³ per hour.Also, we need to figure out the time each pump was working. Since the first and third pumps finished at the same time, let's denote their working time as ( t ) hours. The second pump worked only 2 hours, as stated.Now, let's recall that the amount of water pumped is the product of the rate and time. So:- For the first pump: ( x times t = 9 ) m³. So, ( t = frac{9}{x} ). - For the third pump: ( (x + 3) times t = ) amount pumped by third pump. But we don't know this yet, but we do know that the second and third together pumped 28 m³. Wait, the second pump worked for 2 hours, so its total is ( y times 2 ), and the third pump worked for ( t ) hours, so its total is ( (x + 3) times t ). Together, they pumped 28 m³. So:( 2y + (x + 3)t = 28 ).Also, the combined rate of all three pumps is 14 m³ per hour. So:( x + y + (x + 3) = 14 ).Simplify that: ( 2x + y + 3 = 14 ), so ( 2x + y = 11 ). Let's keep that equation.Now, let's summarize the equations we have:1. From the first pump: ( t = frac{9}{x} ).2. From the second and third pumps: ( 2y + (x + 3)t = 28 ).3. From the combined rate: ( 2x + y = 11 ).Our goal is to find ( x ), ( y ), and ( x + 3 ). Let's substitute ( t ) from equation 1 into equation 2.Substituting ( t = frac{9}{x} ) into equation 2:( 2y + (x + 3)left( frac{9}{x} right) = 28 ).Let's simplify that:First, expand the term with ( (x + 3) ):( 2y + frac{9(x + 3)}{x} = 28 )Which becomes:( 2y + frac{9x + 27}{x} = 28 )Simplify the fraction:( 2y + 9 + frac{27}{x} = 28 )Subtract 9 from both sides:( 2y + frac{27}{x} = 19 )Now, from equation 3, we have ( 2x + y = 11 ). Let's solve equation 3 for ( y ):( y = 11 - 2x )Now, substitute ( y = 11 - 2x ) into the equation ( 2y + frac{27}{x} = 19 ):First, compute 2y:( 2(11 - 2x) = 22 - 4x )So:( 22 - 4x + frac{27}{x} = 19 )Subtract 19 from both sides:( 3 - 4x + frac{27}{x} = 0 )Multiply every term by ( x ) to eliminate the denominator (assuming ( x neq 0 )):( 3x - 4x² + 27 = 0 )Rearranged:( -4x² + 3x + 27 = 0 )Multiply both sides by -1 to make the quadratic coefficient positive:( 4x² - 3x - 27 = 0 )Now, we have a quadratic equation: ( 4x² - 3x - 27 = 0 ). Let's solve for ( x ) using the quadratic formula.Quadratic formula: ( x = frac{-b pm sqrt{b² - 4ac}}{2a} )Here, ( a = 4 ), ( b = -3 ), ( c = -27 ).Compute discriminant:( D = (-3)² - 4*4*(-27) = 9 + 432 = 441 )Square root of 441 is 21.So,( x = frac{-(-3) pm 21}{2*4} = frac{3 pm 21}{8} )So two solutions:1. ( x = frac{3 + 21}{8} = frac{24}{8} = 3 )2. ( x = frac{3 - 21}{8} = frac{-18}{8} = -2.25 )Since a negative pumping rate doesn't make sense, we discard the negative solution. Thus, ( x = 3 ) m³/h.Now, substitute back to find ( y ):From equation 3: ( y = 11 - 2x = 11 - 2*3 = 11 - 6 = 5 ) m³/h.Third pump's rate is ( x + 3 = 3 + 3 = 6 ) m³/h.Let's verify these values to ensure they satisfy all the given conditions.First, check combined rate: 3 + 5 + 6 = 14 m³/h, which matches the given total.Next, check the time ( t ):From first pump: ( t = 9 / 3 = 3 ) hours. So third pump also worked 3 hours, pumping 6 * 3 = 18 m³.Second pump worked 2 hours, pumping 5 * 2 = 10 m³. Together, second and third pumped 10 + 18 = 28 m³, which matches the given.All conditions are satisfied. Therefore, the rates are:First pump: 3 m³/hSecond pump: 5 m³/hThird pump: 6 m³/h

✅:The three pumps have the following pumping rates per hour:- First pump: ( boxed{3 , text{m}^3/text{hour} } )- Second pump: ( boxed{5 , text{m}^3/text{hour} } )- Third pump: ( boxed{6 , text{m}^3/text{hour} } )Step-by-Step Explanation:1. Define Variables: - Let ( x ) be the rate of the first pump (m³/h). - The third pump's rate is ( x + 3 ) m³/h (given). - Let ( y ) be the rate of the second pump (m³/h). - Let ( t ) be the time (hours) the first and third pumps operated (they finished together).2. Key Equations: - First Pump: ( x cdot t = 9 ) → ( t = frac{9}{x} ). - Combined Rate: ( x + y + (x + 3) = 14 ) → ( 2x + y = 11 ). - Second & Third Pumps' Total: ( 2y + (x + 3) cdot t = 28 ).3. Substitute ( t ) into the Third Equation: - Substitute ( t = frac{9}{x} ) into ( 2y + (x + 3) cdot frac{9}{x} = 28 ). - Simplify to: ( 2y + 9 + frac{27}{x} = 28 ) → ( 2y + frac{27}{x} = 19 ).4. Solve for ( y ) from Combined Rate Equation: - ( y = 11 - 2x ).5. Substitute ( y ) into Modified Third Equation: - ( 2(11 - 2x) + frac{27}{x} = 19 ) → ( 22 - 4x + frac{27}{x} = 19 ). - Multiply by ( x ): ( 4x^2 - 3x - 27 = 0 ).6. Solve Quadratic Equation: - Discriminant: ( sqrt{441} = 21 ). - Solutions: ( x = 3 ) (valid) or ( x = -2.25 ) (discarded).7. Find ( y ) and Third Pump's Rate: - ( y = 11 - 2(3) = 5 ). - Third pump: ( 3 + 3 = 6 ).Verification:- Total Rate: ( 3 + 5 + 6 = 14 ) m³/h ✔️.- First Pump: ( 3 times 3 = 9 ) m³ ✔️.- Second & Third Pumps: ( 5 times 2 + 6 times 3 = 10 + 18 = 28 ) m³ ✔️.

🔑:1. Let ( x ) and ( y ) be the amounts of water the first and second pumps can pump per hour, respectively. The problem states that the third pump pumps ( x + 3 ) cubic meters of water per hour. 2. According to the problem, the first pump pumps 9 cubic meters in total, and the second and third pumps together pump 28 cubic meters in total. 3. We can represent the time in which each pump works in terms of the amount of water they pump: - Since the first pump pumps ( x ) cubic meters per hour and it pumps 9 cubic meters in total, it works for ( frac{9}{x} ) hours. - The second pump works for 2 hours in total. - The third pump pumps ( 28 - 2y ). Given it pumps ( x + 3 ) cubic meters per hour, it works for ( frac{28 - 2y}{x+3} ) hours.4. According to the problem, the first and third pumps stop working at the same time. Hence: [ frac{9}{x} = frac{28 - 2y}{x + 3} ]5. Moreover, the three pumps working together for one hour pump 14 cubic meters: [ x + y + (x + 3) = 14 ] Simplifying this, we get: [ 2x + y + 3 = 14 implies 2x + y = 11 ]6. Now we have a system of equations: [ begin{cases} frac{9}{x} = frac{28 - 2y}{x + 3} 2x + y = 11 end{cases} ]7. Solve the second equation for ( y ): [ y = 11 - 2x ]8. Substitute ( y = 11 - 2x ) into the first equation: [ frac{9}{x} = frac{28 - 2(11 - 2x)}{x + 3} ] Simplify inside the numerator: [ frac{9}{x} = frac{28 - 22 + 4x}{x + 3} = frac{6 + 4x}{x + 3} ]9. Cross-multiply to solve for ( x ): [ 9(x + 3) = x(6 + 4x) ] Distribute and combine like terms: [ 9x + 27 = 6x + 4x^2 ] Rearrange into standard form of a quadratic equation: [ 4x^2 - 3x - 27 = 0 ]10. Solve the quadratic equation: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] where ( a = 4 ), ( b = -3 ), and ( c = -27 ): [ x = frac{3 pm sqrt{9 + 432}}{8} = frac{3 pm sqrt{441}}{8} = frac{3 pm 21}{8} ] Thus, ( x ) has two potential values: [ x = frac{24}{8} = 3 quad text{or} quad x = frac{-18}{8} = -2.25 ] Since ( x ) cannot be negative, we have: [ x = 3 ]11. Calculate ( y ) using ( 2x + y = 11 ): [ 2(3) + y = 11 implies 6 + y = 11 implies y = 5 ]12. Finally, calculate the amount of water the third pump can pump per hour: [ x + 3 = 3 + 3 = 6 ]Conclusion:The pumps' rates are:[boxed{3 , text{m}^3 , text{per hour,} , 5 , text{m}^3 , text{per hour, and} , 6 , text{m}^3 , text{per hour.}}]

🧠:Alright, let's tackle these two optimization problems step by step. Both parts (a) and (b) involve finding the smallest possible values of certain expressions given the ranges for x, y, and z. Since these are intervals, I might need to consider the boundaries or maybe some critical points inside the intervals where the minimum could occur. Let me start with part (a).Problem (a): Find the smallest possible value of ( x^{2} + y^{2} ).Okay, so we need to minimize the sum of squares of x and y. Since squares are always non-negative, the smallest possible value would be when both x and y are as close to zero as possible. But let's verify if that's the case here.Given the ranges:- ( x in [-3, 7] )- ( y in [-2, 5] )The square terms ( x^2 ) and ( y^2 ) will be minimized when x and y are closest to zero. Let's check the intervals. For x, the interval goes from -3 to 7. The closest x can get to zero is 0, which is within the interval. Similarly, for y, the interval is from -2 to 5, which also includes 0. Therefore, if we set x = 0 and y = 0, the expression becomes ( 0^2 + 0^2 = 0 ). Wait, but hold on—is 0 within both intervals? Let me check.Yes, x can be 0 because it's between -3 and 7, and y can be 0 because it's between -2 and 5. So the minimal value should be 0. But wait, maybe I'm missing something here. Let me think again. The problem states "the numbers x, y, z are such that..." but z isn't involved in part (a). So maybe z can be ignored here. So, yes, x and y can both be zero. Therefore, the minimal value is 0. Is there any reason that might prevent x and y from being zero? The problem doesn't specify any relationship between x, y, z, so they can vary independently. Hence, yes, the minimal value is indeed 0.But just to be thorough, let's confirm. If x and y can take any values in their intervals independently, then choosing x=0 and y=0 gives the minimal sum. However, sometimes in optimization problems with multiple variables, constraints might tie variables together, but in this case, there are no such constraints mentioned. Therefore, part (a)'s answer should be 0.Problem (b): Find the smallest possible value of ( x y z - z^{2} ).This seems more complex. Let's break it down. We need to minimize the expression ( x y z - z^2 ). Let's note the variables' ranges:- ( x in [-3, 7] )- ( y in [-2, 5] )- ( z in [-5, 3] )So, all three variables are involved here. Let's denote the expression as ( f(x, y, z) = x y z - z^2 ). Our goal is to find the minimum of f(x, y, z) given the intervals for x, y, z.Since this is a function of three variables, the minimum can occur either at critical points inside the domain or on the boundary of the domain. However, because the variables are independent and the function is a polynomial, which is differentiable everywhere, we can check for critical points by setting the partial derivatives to zero. But given that the variables are bounded within intervals, we also need to evaluate the function at the endpoints.But considering the complexity of dealing with three variables, maybe there's a smarter way to approach this. Let me see.First, let's analyze the expression:( f(x, y, z) = z(x y - z) )So, it's factored as z times (xy - z). This factorization might help. Let's see:To minimize f(x, y, z), we can consider the product of z and (xy - z). Since we're dealing with a product, the sign of each term will affect the outcome. To get the smallest possible value (which could be a large negative number), we need to maximize the negative value of the product.Let me consider the two factors: z and (xy - z).Case 1: z is positive. Then, to make the product as negative as possible, (xy - z) should be negative. So, if z is positive, we need (xy - z) < 0 => xy < z. But since z is positive, this would mean that xy is less than a positive number. But depending on x and y, xy can be positive or negative. However, if z is positive, then xy - z could be made more negative by making xy as small as possible (i.e., as negative as possible) and z as large as possible. Wait, but if z is positive and we want (xy - z) to be negative, then making xy as small as possible (most negative) would make (xy - z) even more negative, thus the product z*(xy - z) would be more negative (since z is positive and (xy - z) is negative). Alternatively, if we make z as large as possible, but then (xy - z) becomes more negative only if xy is fixed.Alternatively, perhaps we need to balance between z and (xy - z). This seems a bit tangled. Let's think of specific scenarios.Alternatively, perhaps we can rewrite the expression:( f(x, y, z) = x y z - z^2 = z(x y - z) )Let's consider possible values for z. Since z is in [-5, 3], it can be both positive and negative. Let's analyze the cases where z is positive and where z is negative.Case 1: z ≥ 0 (z ∈ [0, 3])In this case, z is non-negative. Then, the term (xy - z) can be positive or negative depending on xy.To minimize f(x, y, z) = z(xy - z), since z is non-negative, the expression will be minimized when (xy - z) is as negative as possible. Because then, a positive z multiplied by a negative (xy - z) gives a negative number, which could be the minimal value.So, in this case, to make (xy - z) as negative as possible, we need to minimize xy and maximize z. Wait, but z is part of (xy - z), so if z is increased, (xy - z) decreases. But z is multiplied by that term. So perhaps:To minimize f, we can write:For fixed z > 0, the term (xy - z) is minimized when xy is minimized. Therefore, for each z in [0,3], the minimal value of f is z*(min_{x,y} xy - z). So, first find min_{x,y} xy.What's the minimum of xy given x ∈ [-3,7] and y ∈ [-2,5]?The product xy is minimized when either x is negative and y is positive (since that gives a negative product) or x is positive and y is negative (but y can't be less than -2). Wait, y ranges from -2 to 5. So, the minimum of xy occurs when x is as negative as possible and y as positive as possible, or x as positive as possible and y as negative as possible? Let's calculate.The product xy can be minimized (most negative) by taking x as negative as possible and y as positive as possible, or x positive and y negative. Let's compute both possibilities.- x minimum (x = -3), y maximum (y = 5): (-3)(5) = -15- x maximum (x = 7), y minimum (y = -2): 7*(-2) = -14So the minimum of xy is -15, achieved at x = -3, y = 5.Therefore, for each z in [0,3], the minimal value of (xy - z) is (-15 - z). Therefore, the minimal f in this case would be z*(-15 - z) = -15z - z². Now, to find the minimum of this expression over z ∈ [0,3].Wait, but hold on. Wait, for a given z, we fix z and then find the minimal xy. But if we fix z, and then we can choose x and y to minimize xy. But if we can choose x and y independently, then regardless of z, the minimal xy is always -15. But actually, when we fix z, we can still choose x and y to minimize xy. So indeed, the minimal (xy - z) is (-15 - z). Therefore, f(z) = z*(-15 - z) = -15z - z². Now, we need to find the minimal value of this quadratic function over z ∈ [0,3].But wait, actually, we are looking for the minimum of f(x,y,z) over all x,y,z in their intervals. However, in this case, for each z, the minimal f is z*(-15 - z). So, to find the overall minimum, we need to find the minimal value of -15z - z² for z in [0,3].Let's analyze the function g(z) = -15z - z² over z ∈ [0,3]. This is a downward-opening parabola (since the coefficient of z² is negative) but we're looking at it over an interval. The vertex of the parabola is at z = -b/(2a) where a = -1, b = -15. So, the vertex is at z = -(-15)/(2*(-1)) = 15/(-2) = -7.5. But this is outside our interval [0,3]. Therefore, over [0,3], the function g(z) is decreasing because the derivative g’(z) = -15 - 2z, which is always negative (since z ≥ 0, so -15 -2z ≤ -15 < 0). Therefore, the function is decreasing on [0,3]. Hence, its minimum occurs at the right endpoint z = 3. So, the minimal value in this case is g(3) = -15*3 - 3² = -45 -9 = -54.Case 2: z < 0 (z ∈ [-5, 0))Now, when z is negative, the expression f(x, y, z) = z*(xy - z). Since z is negative, to minimize the expression (which is a product of z and (xy - z)), we need to make (xy - z) as positive as possible. Because a negative number multiplied by a positive number gives a negative number, and to get the smallest possible value (most negative), we need the product to be as negative as possible. However, since z is negative, making (xy - z) positive would result in a negative product. But how can we maximize the positivity of (xy - z) when z is negative?Wait, let's clarify. If z is negative, then (xy - z) becomes xy - z. Since z is negative, subtracting a negative is like adding a positive. So, (xy - z) = xy + |z|. To make this term as large as possible (positive), since we are multiplying by z (which is negative), the larger (xy - z) is, the more negative the product becomes. Therefore, to minimize f(x,y,z), when z is negative, we need to maximize (xy - z). Since z is negative, this is equivalent to maximizing (xy + |z|). To maximize (xy + |z|), we need to maximize xy and |z|.But since |z| is maximized when z is most negative (i.e., z = -5). However, for each fixed z, to maximize (xy - z), we need to maximize xy. Let's formalize this.For z < 0, f(x,y,z) = z*(xy - z). Since z is negative, and we want to minimize f, which is equivalent to making z*(xy - z) as small as possible (most negative). To do this, since z is negative, we need (xy - z) to be as large as possible (positive). Because a negative times a positive is negative, and larger positive makes the product more negative.Thus, for each fixed z < 0, the maximum of (xy - z) is (max_{x,y} xy) - z. So first, we need to find the maximum of xy over x ∈ [-3,7], y ∈ [-2,5].To find the maximum of xy, similar to before. The maximum occurs when both x and y are positive and as large as possible, or one is negative and the other is negative. Let's compute:- x maximum (7), y maximum (5): 7*5=35- x minimum (-3), y minimum (-2): (-3)*(-2)=6So the maximum of xy is 35, achieved at x=7, y=5.Therefore, for each fixed z < 0, the maximum of (xy - z) is 35 - z. Hence, f(z) = z*(35 - z). So, we have to find the minimum of this function over z ∈ [-5, 0).Let's analyze the function h(z) = z*(35 - z) = 35z - z². This is a quadratic function in z. Let's find its behavior over z ∈ [-5, 0).The function h(z) = -z² + 35z. Since the coefficient of z² is negative, it's a downward-opening parabola. The vertex is at z = -b/(2a) = -35/(2*(-1)) = 35/2 = 17.5, which is way outside our interval [-5, 0). Therefore, on the interval [-5, 0), the function h(z) is increasing because the derivative h’(z) = 35 - 2z. Since z ≤ 0, h’(z) = 35 - 2z ≥ 35 - 0 = 35 > 0. Thus, the function is increasing on this interval. Therefore, the minimum occurs at the left endpoint z = -5.Calculating h(-5) = (-5)*(35 - (-5)) = (-5)*(40) = -200.Wait, that's a very large negative number. That seems like a candidate for the minimal value. But let's confirm.Wait, when z = -5, the expression f(x,y,z) = z*(xy - z). To achieve the maximum (xy - z) when z = -5, we set x=7 and y=5 (to maximize xy), so (xy - z) = 35 - (-5) = 40. Then f = (-5)*40 = -200. So, that's possible. Let's check if this is indeed attainable. x=7 is within [-3,7], y=5 is within [-2,5], z=-5 is within [-5,3]. Yes, all values are within their respective intervals. Therefore, this is a valid point.Now, comparing the two cases:- Case 1 (z ≥ 0): minimal value is -54 at z=3, x=-3, y=5- Case 2 (z < 0): minimal value is -200 at z=-5, x=7, y=5Therefore, the overall minimal value is -200.But wait a second, let me double-check. Because in Case 1, when z=3, x=-3, y=5. Then f = 3*(-15 -3) = 3*(-18) = -54. That seems correct.In Case 2, when z=-5, x=7, y=5. Then f = -5*(35 - (-5)) = -5*40 = -200. That's correct.But is there any other case where we might get a smaller value? Let me think. For example, if instead of maximizing xy in Case 2, could we get a more negative product by choosing different x and y?Wait, in Case 2, we are trying to maximize (xy - z) because z is negative, so we want (xy - z) to be as large as possible. Since z is negative, (xy - z) = xy + |z|. Therefore, the maximum of (xy + |z|) occurs when xy is maximum and |z| is maximum. However, in our approach, for each z, we took the maximum xy (35) and then considered the effect over z. But actually, for a fixed z, |z| is fixed, so maximizing xy would indeed maximize (xy + |z|). But since z can vary, perhaps there's a better combination where both xy is large and |z| is large.Wait, but in our analysis, we considered for each z, the maximum xy is 35, so (xy - z) = 35 - z. Then, h(z) = z*(35 - z). So, if z is more negative (i.e., z = -5), then (35 - z) becomes 35 - (-5) = 40, so h(z) = -5*40 = -200. If z is less negative, say z = -4, then h(z) = -4*(35 - (-4)) = -4*39 = -156, which is less negative than -200. Therefore, indeed, the minimum occurs at z = -5. Therefore, the minimal value is -200.But let's check another angle. Suppose we fix x and y to something else. For example, suppose z is negative, but we take x and y such that xy is not maximum. Could that result in a lower f?Wait, suppose instead of taking x=7 and y=5 (max xy), we take x=7 and y=-2 (min y). Then, xy = 7*(-2) = -14. Then, (xy - z) = -14 - z. Since z is negative, say z = -5, then (xy - z) = -14 - (-5) = -14 +5 = -9. Then f = z*(-9) = (-5)*(-9) = 45, which is positive. That's not helpful for minimizing.Alternatively, take x=-3 and y=5 (which gives xy=-15). Then, (xy - z) = -15 - z. If z is negative, say z=-5, then (xy - z) = -15 - (-5) = -10. Then f = z*(-10) = (-5)*(-10) = 50, which is positive. Again, not helpful.Alternatively, take x=-3 and y=-2, xy=6. Then, (xy - z)=6 - z. If z=-5, this is 6 - (-5)=11. Then f = (-5)*11 = -55. That's better than -54 but worse than -200.Alternatively, take x=7, y=5, z=-5: that gives us -200 as before. So indeed, the minimal value is achieved when xy is maximized (35) and z is minimized (-5). So, that seems to be the case.Is there another combination where maybe z is not at its minimum, but somewhere else, but the product is more negative? Let's suppose z=-5, x=7, y=5: f=-200. If we take z=-4, then to maximize (xy - z), we need xy=35, so (35 - (-4))=39, then f=(-4)*39=-156, which is higher (less negative). Similarly, for z=-3: f=(-3)*(35 - (-3))= -3*38= -114. So, as z increases towards 0, the value becomes less negative. Therefore, the minimal value is indeed at z=-5.Alternatively, what if we take a lower z? But z can't be lower than -5. So, that's the minimum z.Wait, another thought: Could we get a lower value by having (xy - z) negative when z is negative? Because then, z*(negative) would be positive. But since we're trying to minimize, positive values are worse. So, no. Therefore, the strategy of making (xy - z) as large as possible when z is negative is correct.Therefore, the minimal value of ( x y z - z^2 ) is -200.But let me check another possibility. Suppose instead of factoring, we consider the original expression: ( x y z - z^2 ). Maybe we can complete the square or something. Let's see.( x y z - z^2 = z(x y - z) ). Not sure if that helps, but maybe think of it as quadratic in z: ( -z^2 + x y z ). For fixed x and y, this is a quadratic in z, which opens downward (since the coefficient of z² is -1). The maximum occurs at z = (x y)/2. But since we're looking for the minimum over z in [-5,3], given that the quadratic opens downward, the minimum would occur at one of the endpoints, either z=-5 or z=3, depending on x and y.But since x and y can vary, perhaps we can choose x and y such that when z is at an endpoint, the expression is minimized.Wait, for each x and y, the expression as a function of z is a quadratic with maximum at z=(xy)/2. So, if we fix x and y, then the minimal value of the expression over z ∈ [-5,3] would be the minimum between z=-5 and z=3. Therefore, for each x and y, the minimal of the expression is the minimum between:At z=-5: -5*(x y) - (-5)^2 = -5 x y -25At z=3: 3 x y -9Therefore, the minimal value of the entire expression is the minimum over all x ∈ [-3,7], y ∈ [-2,5] of the minimum between (-5 x y -25) and (3 x y -9). So, perhaps we can compute the minimal value among these two expressions.But this seems complicated. Let me see.Alternatively, since we have to consider all x, y, z, and for each x, y, the minimal over z is the minimum of (-5xy -25) and (3xy -9). Therefore, the overall minimal value is the minimum over all x, y of min(-5xy -25, 3xy -9).But this might not capture the actual minimal, because for some x and y, the minimal could be at z=-5, and for others at z=3. However, we need to find the global minimum over all x, y, z.But maybe we can compare these two expressions.Let me denote A = -5xy -25 (when z=-5)and B = 3xy -9 (when z=3)We need to find min_{x,y} [min(A, B)]To find the overall minimum, we need to see which of A or B can attain the lower value.Suppose for some x and y, A < B. Then the minimal is A. For others, if B < A, then minimal is B. We need to find the smallest among all such minima.But this approach might be tricky. Let's see if we can find x and y such that A is minimized and B is as low as possible.Alternatively, perhaps the minimal value occurs when A is minimized or when B is minimized, whichever is smaller.Let's compute the minimal value of A and the minimal value of B.First, minimal value of A = -5xy -25.To minimize A, we need to maximize xy, because it's multiplied by -5. The maximum of xy is 35 (as before, x=7, y=5). Therefore, minimal A = -5*35 -25 = -175 -25 = -200.Similarly, minimal value of B = 3xy -9.To minimize B, we need to minimize xy. The minimal xy is -15 (x=-3, y=5). Therefore, minimal B = 3*(-15) -9 = -45 -9 = -54.Therefore, the overall minimal value is the minimum between -200 and -54, which is -200.Therefore, this confirms our previous result. The minimal value is -200, achieved when A is minimized, which corresponds to x=7, y=5, z=-5.Therefore, after analyzing both cases and verifying through different methods, it seems that the minimal value for part (b) is indeed -200.Summary of Findings:- For part (a), since both x and y can reach zero within their intervals, the minimal value of ( x^2 + y^2 ) is 0.- For part (b), through analyzing the expression by considering the sign of z and optimizing accordingly, the minimal value is found to be -200 when x=7, y=5, and z=-5.

✅:(a) The smallest possible value of ( x^2 + y^2 ) is achieved when both ( x ) and ( y ) are as close to zero as possible. Since ( x in [-3, 7] ) and ( y in [-2, 5] ), both variables can independently attain 0. Thus, the minimum value is ( 0^2 + 0^2 = 0 ).Answer: (boxed{0})(b) To minimize ( xyz - z^2 ), we factor it as ( z(xy - z) ). Analyzing cases for ( z geq 0 ) and ( z < 0 ):- Case 1 (( z geq 0 )): Minimize ( z(xy - z) ). For fixed ( z ), minimize ( xy ) (achieved at ( x = -3, y = 5 ), giving ( xy = -15 )). The expression becomes ( -15z - z^2 ), minimized at ( z = 3 ), yielding ( -54 ).- Case 2 (( z < 0 )): Maximize ( xy - z ) (equivalent to maximizing ( xy ), achieved at ( x = 7, y = 5 ), giving ( xy = 35 )). The expression becomes ( z(35 - z) ), minimized at ( z = -5 ), yielding ( -200 ).Comparing both cases, the smallest value is (-200) when ( x = 7 ), ( y = 5 ), and ( z = -5 ).Answer: (boxed{-200})

🔑:# Part (a)1. Given the ranges of the variables: [ x in [-3, 7], quad y in [-2, 5] ]2. Our goal is to find the minimum possible value of ( x^2 + y^2 ).3. Note that both ( x^2 geq 0 ) and ( y^2 geq 0 ) because squaring any real number results in a non-negative value.4. Therefore, [ x^2 + y^2 geq 0 ]5. To achieve ( x^2 + y^2 = 0 ), we need both ( x = 0 ) and ( y = 0 ). These values are within the given ranges: [ x = 0 subseteq [-3, 7] quad text{and} quad y = 0 subseteq [-2, 5] ]6. Hence, the minimum value is when ( x = 0 ) and ( y = 0 ): [ x^2 + y^2 = 0^2 + 0^2 = 0 ]Conclusion: [ boxed{0} ]# Part (b)1. Given the ranges of the variables: [ x in [-3, 7], quad y in [-2, 5], quad z in [-5, 3] ]2. Our goal is to find the minimum possible value of ( xy z - z^2 ).3. Using the absolute values, we can estimate the maximum value: [ |xyz| = |x| cdot |y| cdot |z| leq 7 cdot 5 cdot 5 = 175 ] Similarly, [ z^2 leq 5^2 = 25 ]4. So, [ xy z - z^2 geq -|xyz| - z^2 = -175 - 25 = -200 ]5. Now, check if this minimum value can be attained. Set ( x = 7 ), ( y = 5 ), and ( z = -5 ): [ x y z - z^2 = 7 cdot 5 cdot (-5) - (-5)^2 ] [ = 7 cdot 5 cdot (-5) - 25 ] [ = -175 - 25 ] [ = -200 ]Conclusion:[ boxed{-200} ]

🧠:Okay, let's see. I need to find the value of (3x + y + z)/y given that x/3 = y/(-4) = z/7. Hmm, okay. So the problem gives me a proportion involving x, y, and z, and I need to compute this expression. Let me think step by step.First, when I see that x/3 = y/(-4) = z/7, this seems like a set of equal ratios. So maybe I can set each of these equal to a common variable, like k. That way, I can express x, y, and z in terms of k. Let me try that.Let’s let k = x/3 = y/(-4) = z/7. Then, from this, I can write:x = 3k,y = -4k,z = 7k.Yes, that seems right. So substituting each ratio into k gives expressions for x, y, z in terms of k. Now, I can substitute these expressions into the numerator 3x + y + z and then divide by y.Let me compute the numerator first. 3x would be 3*(3k) = 9k. Then y is -4k, and z is 7k. So adding them together: 9k + (-4k) + 7k. Let me calculate that.9k -4k is 5k, and 5k +7k is 12k. So the numerator is 12k. The denominator is y, which is -4k. So the entire expression becomes 12k / (-4k). Wait, but here, k is a common factor, so as long as k ≠ 0, we can simplify this. So 12k divided by -4k is equal to (12/-4)*(k/k) = -3*1 = -3. So the value is -3. So the answer should be (D) -3.But wait, let me check my steps again to make sure I didn't make a mistake. Let me verify each part.First, setting k = x/3 = y/(-4) = z/7. Then solving for x, y, z:x = 3k,y = -4k,z = 7k.Yes, that's correct. Then substituting into 3x + y + z:3x = 3*(3k) = 9k,y = -4k,z =7k,Adding them: 9k -4k +7k. Let's compute term by term:9k -4k is indeed 5k. Then 5k +7k is 12k. So numerator is 12k. Denominator is y = -4k.So 12k / (-4k). Since k is not zero (because if k were zero, then x, y, z would all be zero, and then y would be zero, which would make the expression undefined. So k can't be zero, so we can cancel out k. So 12/-4 = -3. So yes, the value is -3. Therefore, the correct answer is (D) -3.Wait, but let me think if there's another way to approach this problem, maybe without introducing a variable k. For example, using ratios directly.Given that x/3 = y/(-4) = z/7, this implies that x, y, z are proportional to 3, -4, 7. So maybe we can write x = 3k, y = -4k, z =7k, which is exactly what I did before. So the same result.Alternatively, maybe express x and z in terms of y. Let's see. Since x/3 = y/(-4), so x = (3/(-4))y. Similarly, z/7 = y/(-4), so z = (7/(-4))y. Then substitute these into the expression (3x + y + z)/y.Let me try this approach.Express x and z in terms of y:x = (3/(-4))y = (-3/4)y,z = (7/(-4))y = (-7/4)y.Then compute 3x + y + z:3x = 3*(-3/4)y = (-9/4)y,y remains as y,z = (-7/4)y.Adding them together: (-9/4)y + y + (-7/4)y.Convert y to 4/4 y to combine fractions:(-9/4)y + (4/4)y + (-7/4)y = [(-9 +4 -7)/4]y = (-12/4)y = -3y.So numerator is -3y, denominator is y, so (-3y)/y = -3. Since y ≠ 0 (because if y is zero, then x and z would also be zero, making the original ratios undefined), so we can safely divide by y. Thus, the value is -3. So again, answer is (D) -3.Therefore, both methods lead to the same conclusion, which is reassuring. Let me check once more if there's any possible mistake.Another way: suppose we pick a specific value for k. Let's choose k =1. Then x=3, y=-4, z=7. Plug into the expression:3x + y + z = 3*3 + (-4) +7 = 9 -4 +7 = 12. Then 12 / (-4) = -3. So same result. If we take k=2, x=6, y=-8, z=14. Then 3x + y + z =18 -8 +14=24. Then 24 / (-8)= -3. Same answer. Therefore, regardless of the value of k, as long as k≠0, the ratio is -3.Thus, the answer is definitely (D) -3.I think that's thorough. I considered multiple approaches, checked with substitution, verified with different methods, and all lead to the same answer. So I can be confident that (D) is correct.

✅:Given the proportion ( frac{x}{3} = frac{y}{-4} = frac{z}{7} ), we introduce a common variable ( k ) such that:[x = 3k, quad y = -4k, quad z = 7k.]Substituting these into the expression ( frac{3x + y + z}{y} ):[3x = 3(3k) = 9k, quad y = -4k, quad z = 7k.]Adding the terms in the numerator:[3x + y + z = 9k - 4k + 7k = 12k.]The denominator is ( y = -4k ), so:[frac{12k}{-4k} = frac{12}{-4} = -3.]Verification using direct substitution (e.g., ( k = 1 )) confirms the result. Thus, the value is (boxed{D}).

🔑:Given that ( frac{x}{3} = frac{y}{-4} = frac{z}{7} ), let's denote this common ratio by ( k ). Therefore, we have:1. ( x = 3k )2. ( y = -4k )3. ( z = 7k )Our goal is to find the value of ( frac{3x + y + z}{y} ).1. Substitute the expressions for ( x ), ( y ), and ( z ) into the equation:[ frac{3x + y + z}{y} = frac{3(3k) + (-4k) + 7k}{-4k} ]2. Simplify the numerator:[3(3k) + (-4k) + 7k = 9k - 4k + 7k ]3. Adding the terms in the numerator:[9k - 4k + 7k = 12k ]4. The expression now becomes:[frac{12k}{-4k}]5. Simplify the fraction by canceling ( k ):[frac{12k}{-4k} = frac{12}{-4} = -3]Conclusion: Thus, the value of ( frac{3x + y + z}{y} ) is ( boxed{D} ).

🧠:Alright, let's try to tackle this problem. The question says that if we have two integers a and b, and the sum of their greatest common divisor (gcd) and least common multiple (lcm) equals the sum of the numbers themselves, then one of the numbers must divide the other. So, we need to show that either a divides b or b divides a given that gcd(a, b) + lcm(a, b) = a + b.First, let me recall some properties of gcd and lcm. I remember that for any two integers a and b, the product of the gcd and lcm of a and b is equal to the product of a and b. So, mathematically, that's:gcd(a, b) × lcm(a, b) = a × bThis is a crucial formula. Maybe this can be helpful here. Let's denote d = gcd(a, b). Then, we can express a and b in terms of d. Let me write that down.Let d = gcd(a, b). Then, we can write a = d × m and b = d × n, where m and n are integers that are coprime, meaning gcd(m, n) = 1. This is a standard way to represent a and b when dealing with gcd and lcm.Given that, the lcm(a, b) would be d × m × n, since the lcm of two numbers is the product divided by the gcd. Wait, actually, since a = d × m and b = d × n, then lcm(a, b) is d × m × n because m and n are coprime. Let me confirm that. The general formula is lcm(a, b) = (a × b) / gcd(a, b). Substituting a and b:lcm(a, b) = (d × m × d × n) / d = d × m × n. Yes, that's correct.So, now, let's write down the equation given in the problem:gcd(a, b) + lcm(a, b) = a + bSubstituting the expressions in terms of d, m, and n:d + (d × m × n) = d × m + d × nLet me factor out the d from all terms:d × [1 + m × n] = d × (m + n)Since d is not zero (because a and b are integers, and gcd is at least 1 if both are non-zero; but if either a or b is zero, the gcd is the absolute value of the other. Wait, actually, the problem states a and b are integers, but they could be zero. Wait, but gcd(0, b) is |b|, right? Hmm, but if a and b are both zero, then gcd is zero and lcm is undefined. So maybe the problem assumes a and b are positive integers. Let me check.Wait, the original problem says "Let a, b ∈ ℤ", so integers, which includes zero and negative numbers. However, gcd is always non-negative, and lcm is also non-negative. The problem doesn't specify positivity, but perhaps we can assume they are positive? Because if one of them is zero, say a = 0, then gcd(0, b) = |b|, and lcm(0, b) is 0, so the equation would be |b| + 0 = 0 + b. If b is positive, this is b + 0 = 0 + b, which is true, but then a divides b (since 0 divides b only if b is 0, but b can be any integer here). Wait, actually, if a = 0 and b ≠ 0, then 0 does not divide b. Hmm, maybe there's an issue here. Let me think.Wait, but the problem states "a, b ∈ ℤ", but maybe we need to assume they are positive integers. Otherwise, the problem might have exceptions. Let me check the original problem again. It says "Show that a ∣ b or b ∣ a". So, for example, if a = 0 and b ≠ 0, then a does not divide b, unless b is 0. But 0 divides only 0. So if a = 0 and b ≠ 0, then 0 does not divide b, and unless b divides 0, which is always true because any integer divides 0. So if a = 0 and b ≠ 0, then b divides a (since 0 is divisible by any integer). Wait, but the problem says "a divides b or b divides a". So in the case where a = 0 and b ≠ 0, we have b divides a (since a is zero). So in that case, the conclusion holds. If both a and b are zero, then both divide each other, trivially. So maybe the problem holds even if a or b is zero. Let's verify.Suppose a = 0 and b ≠ 0. Then gcd(0, b) = |b|, lcm(0, b) = 0. Then the equation is |b| + 0 = 0 + b. If b is positive, then |b| = b, so this becomes b = b, which is true. If b is negative, then |b| = -b, so the left side is -b + 0 = -b, and the right side is 0 + b = b. So for negative b, we have -b = b, which implies b = 0, but we assumed b ≠ 0. Therefore, in the case a = 0 and b ≠ 0, the equation holds only if b is positive. But in integers, if a = 0 and b is negative, then the equation would not hold. However, since gcd is non-negative, and lcm is non-negative, but a and b can be negative. Hmm, this complicates things.Alternatively, maybe the problem is intended for positive integers. Let me assume a and b are positive integers. That way, we don't have to deal with negative numbers or zero. So, proceeding under that assumption, a and b are positive integers, which is standard in number theory problems unless otherwise specified.So, back to the equation. We have d(1 + mn) = d(m + n). Since d is a positive integer, we can divide both sides by d, yielding:1 + mn = m + nSo now the equation simplifies to mn - m - n + 1 = 0. Let me rearrange that:mn - m - n + 1 = 0This looks like it can be factored. Let me see:mn - m - n + 1 = (m - 1)(n - 1) = 0Yes, because (m - 1)(n - 1) = mn - m - n + 1. So the equation becomes:(m - 1)(n - 1) = 0Therefore, either m - 1 = 0 or n - 1 = 0. Which implies that either m = 1 or n = 1.Recall that m and n are coprime, as we set a = d*m and b = d*n with gcd(m, n) = 1. But since they are coprime, and one of them is 1, then:If m = 1, then a = d*1 = d, and b = d*n. Since m = 1 and n is an integer, but gcd(m, n) = 1 implies n can be any integer, but since m = 1, n can be any integer coprime to 1, which is any integer. But since a and b are positive, n must be a positive integer. So, if m = 1, then a = d, and b = d*n. Therefore, a divides b, because b = a*n. Similarly, if n = 1, then b = d*1 = d, and a = d*m, so b divides a, since a = b*m. Therefore, in either case, one divides the other. Therefore, the conclusion holds.Therefore, under the assumption that a and b are positive integers, we have shown that a divides b or b divides a.But wait, let me check if we need to consider the case where a or b is zero. Let's see. If a = 0 and b ≠ 0, then as we saw earlier, gcd(0, b) = |b|, lcm(0, b) = 0, so the equation becomes |b| + 0 = 0 + b. This holds if b is positive, but not if b is negative. So if a = 0 and b is positive, then the equation holds, and the conclusion is that b divides a (since a = 0), which is true because any number divides 0. If b is negative, the equation does not hold. However, if we allow a and b to be negative, we might have issues.Alternatively, if we take absolute values into account. Wait, but gcd is always positive, lcm is also always positive. But a and b can be negative. Let me test with a negative example.Suppose a = -2 and b = 4. Then gcd(-2, 4) is 2, lcm(-2, 4) is 4. Then gcd + lcm = 2 + 4 = 6. The sum a + b = -2 + 4 = 2. 6 ≠ 2, so the equation doesn't hold. So in this case, the equation is not satisfied. What if a and b are such that one divides the other?Take a = -2 and b = -4. Then gcd(-2, -4) is 2, lcm(-2, -4) is 4. Then gcd + lcm = 2 + 4 = 6. The sum a + b = -2 + (-4) = -6. 6 ≠ -6, so the equation does not hold. But here, a divides b (since -4 divided by -2 is 2). So even though a divides b, the equation doesn't hold because of the signs.Alternatively, if we take a = 2 and b = -2. Then gcd(2, -2) = 2, lcm(2, -2) = 2. Then gcd + lcm = 2 + 2 = 4. Sum a + b = 2 + (-2) = 0. 4 ≠ 0. So equation doesn't hold. Hmm. So maybe the problem implicitly assumes that a and b are positive integers. Because with negative integers, even if one divides the other, the equation might not hold.Alternatively, maybe we need to take absolute values in the equation. Wait, but the problem states gcd(a, b) + lcm(a, b) = a + b. So if a and b are negative, their sum is negative, but gcd and lcm are positive, so the left side is positive, and the right side is negative or positive depending on a and b. So unless a and b are both positive, the equation may not hold. Therefore, perhaps the problem is intended for positive integers a and b. Therefore, we can safely proceed under that assumption, and the conclusion follows as shown earlier.Therefore, the key steps are:1. Let d = gcd(a, b), then write a = d*m, b = d*n with gcd(m, n) = 1.2. Substitute into the equation gcd(a, b) + lcm(a, b) = a + b to get d + d*m*n = d*m + d*n.3. Divide both sides by d to obtain 1 + m*n = m + n.4. Rearrange to (m - 1)(n - 1) = 0, which implies m = 1 or n = 1.5. Therefore, if m = 1, then a = d divides b = d*n; if n = 1, then b = d divides a = d*m.Hence, the conclusion follows.But wait, let me test with an example where a divides b. Take a = 2 and b = 4. Then gcd(2, 4) = 2, lcm(2, 4) = 4. Then gcd + lcm = 2 + 4 = 6, and a + b = 6. So equation holds. Similarly, if a divides b, it works.Another example where b divides a. Let a = 6 and b = 3. Then gcd(6, 3) = 3, lcm(6, 3) = 6. Then 3 + 6 = 9, and a + b = 9. So equation holds.What if neither divides the other? Let's pick a = 2 and b = 3. Then gcd(2, 3) = 1, lcm(2, 3) = 6. Then 1 + 6 = 7, but a + b = 5. 7 ≠ 5, so equation doesn't hold. So indeed, if neither divides the other, the equation does not hold. Therefore, the converse is also true: if the equation holds, then one must divide the other.Another example: a = 3 and b = 6. Then same as above, works. What about a = 4 and b = 6. Then gcd(4, 6) = 2, lcm(4, 6) = 12. Then 2 + 12 = 14, and a + b = 10. 14 ≠ 10. So equation doesn't hold, which aligns with the conclusion since neither 4 divides 6 nor 6 divides 4.Another case: a = 1 and b = 1. Then gcd(1, 1) = 1, lcm(1, 1) = 1. So 1 + 1 = 2, and a + b = 2. Equation holds. Here, each divides the other.So all test cases seem to support the conclusion. Therefore, the reasoning seems solid.Therefore, the answer is that one of a or b must divide the other if gcd(a, b) + lcm(a, b) = a + b.Final Answerboxed{a mid b text{ or } b mid a}

✅:Given integers (a) and (b) such that (gcd(a, b) + operatorname{lcm}(a, b) = a + b), we need to show that either (a mid b) or (b mid a).1. Express (a) and (b) in terms of their gcd: Let (d = gcd(a, b)). Then we can write (a = d cdot m) and (b = d cdot n) where (gcd(m, n) = 1).2. Substitute into the given equation: [ gcd(a, b) + operatorname{lcm}(a, b) = a + b ] Substituting (a = d cdot m) and (b = d cdot n), we get: [ d + d cdot m cdot n = d cdot m + d cdot n ]3. Simplify the equation: Divide both sides by (d) (since (d neq 0)): [ 1 + m cdot n = m + n ] Rearrange the terms: [ m cdot n - m - n + 1 = 0 ] Factor the left-hand side: [ (m - 1)(n - 1) = 0 ]4. Solve the factored equation: The product ((m - 1)(n - 1) = 0) implies (m = 1) or (n = 1).5. Conclude the divisibility: - If (m = 1), then (a = d cdot 1 = d) and (b = d cdot n). Hence, (a mid b). - If (n = 1), then (b = d cdot 1 = d) and (a = d cdot m). Hence, (b mid a).Thus, we have shown that either (a mid b) or (b mid a).[boxed{a mid b text{ or } b mid a}]

🔑:1. Soient ( a, b in mathbb{Z} ) tels que (operatorname{pgcd}(a, b) + operatorname{ppcm}(a, b) = a + b ). 2. Définissons (delta = operatorname{pgcd}(a, b) ). 3. Remarquons que (operatorname{ppcm}(a, b) = frac{a cdot b}{operatorname{pgcd}(a, b)} = frac{a cdot b}{delta}).4. En substituant ces valeurs dans l'équation initiale, nous obtenons (delta + frac{a cdot b}{delta} = a + b).5. Multiplions chaque terme de l'équation par (delta) pour éliminer le dénominateur:[delta^2 + a cdot b = delta cdot a + delta cdot b.]6. Simplifions cette équation:[delta^2 + a cdot b - delta cdot a - delta cdot b = 0.]À savoir,[delta^2 + a cdot b = delta cdot a + delta cdot b.]7. Réorganisions les termes pour faciliter la factorisation:[delta^2 - delta cdot a - delta cdot b + a cdot b = 0.]8. Factorisons les termes:[delta^2 - delta (a+b) + ab = 0,]ce qui peut être réécrit comme:[(delta - a)(delta - b) = 0.]9. Pour que le produit de ces deux termes soit zéro, soit ((delta - a) = 0) soit ((delta - b) = 0).10. Par conséquent, soit (delta = a ) soit (delta = b ), donc:[operatorname{pgcd}(a, b) = a text{ ou } operatorname{pgcd}(a, b) = b.]11. Si (operatorname{pgcd}(a, b) = a), alors ( a mid b). Si (operatorname{pgcd}(a, b) = b), alors ( b mid a). Conclusion:Nous avons montré que (operatorname{pgcd}(a, b) = a ) ou (operatorname{pgcd}(a, b) = b), ce qui implique que ( a mid b ) ou ( b mid a ).[boxed{a mid b text{ ou } b mid a}]