Can you solve this Calculus optimization problem?

Let x > 0, y > 0 xy = 192 z = x + y <-- function to be minimized y = 192/x z = x + (192/x) z' = 1 - (192/x²) = 0 x² = 192 x = √192 = 8√3 y = 192/8√3 = 24/√3 = 8√3 x = y = 8√3 This is basically the same as having a set area for a rectangle and finding the minimum perimeter (or semi-perimeter) so it shouldn't be a surprise that x = y.

Let x > 0, y > 0 xy = 192 z = x + y <-- function to be minimized y = 192/x z = x + (192/x) z' = 1 - (192/x²) = 0 x² = 192 x = √192 = 8√3 y = 192/8√3 = 24/√3 = 8√3 x = y = 8√3 This is basically the same as having a set area for a rectangle and finding the minimum perimeter (or semi-perimeter) so it shouldn't be a surprise that x = y.

sqrt(192)

Call the numbers x and y. Call the product P. Our condition is that P = x*y The sum is: S = x + y Replace y with P/x: S = x + P/x This is our objective equation. Take its derivative and set equal to zero, as is the procedure in optimization. dS/dx = 0 dS/dx = 1 - P/x^2 1 - P/x^2 = 0 Thus: 1 = P/x^2 x^2 = P Thus, x = sqrt(P) Corresponding value of y: y = P/sqrt(P) y = sqrt(P) x = sqrt(P) As a conclusion with P:=192: x = 8*sqrt(3) = 13.856 y = 8*sqrt(3) = 13.856 If you must have a pair of integers: x = 12 y = 16 or vice-versa

i guess you dont need calculus to solve this very simple problem. i suppose you need not just positive number but positive whole number 192 is 2^6 *3 by prime factorization distribute the exponents as even as possible and try different factors and get the sum you get 12 and 16. if you do the calculus, youll get x^2 = 192 or x= ~ 13.85 get factors close to this value, you get 12 and 16.

Call the two numbers x and y. Since their product is 192, we can write xy=192, or y=192/x. We need to minimize the sum of the two numbers, which is equivalent to minimizing the function S(x,y)=x+y. Since we have y in terms of x, we can rewrite that function as S(x)=x+192/x. To minimize that function, we have to find where the derivative is zero. So first find the derivative: S(x)=x+192/x S(x) = x + 192x^-1 S'(x) = 1 - 192x^-2 S'(x) = 1 - 192/x^2 Now set the derivative equal to zero and solve for x: 1 - 192/x^2 = 0 192/x^2 = 1 x^2 = 192 x = ±sqrt(192) x = sqrt(192) (The numbers are required to be positive). x = sqrt(64*3) x = 8*sqrt(3) Now solve for y: y = 192/x y = 192/[8*sqrt(3)] y = 24/sqrt(3) y = 24*sqrt(3)/3 y = 8*sqrt(3) So both numbers are 8*sqrt(3). Note that there was no requirement that the two numbers be distinct. I hope that helps :)

this actually more of a common sense problem. let's say that we change the question to 144 instead then we ask is 144+1 > 12 + 12. Yes! so we know that numbers that are close to eachother have a sum that is smaller. so.... 192 = 2^6 * 3 = 3*2^2 * 2^4 = 16 * 12 this is ofcorse assuming that we are only doing integers. If not then sqrt(192) * sqrt(192) would be the smallest sum

Call the two numbers x and y. Since their product is 192, we can write xy=192, or y=192/x. We need to minimize the sum of the two numbers, which is equivalent to minimizing the function S(x,y)=x+y. Since we have y in terms of x, we can rewrite that function as S(x)=x+192/x. To minimize that function, we have to find where the derivative is zero. So first find the derivative: S(x)=x+192/x S(x) = x + 192x^-1 S'(x) = 1 - 192x^-2 S'(x) = 1 - 192/x^2 Now set the derivative equal to zero and solve for x: 1 - 192/x^2 = 0 192/x^2 = 1 x^2 = 192 x = ±sqrt(192) x = sqrt(192) (The numbers are required to be positive). x = sqrt(64*3) x = 8*sqrt(3) Now solve for y: y = 192/x y = 192/[8*sqrt(3)] y = 24/sqrt(3) y = 24*sqrt(3)/3 y = 8*sqrt(3) So both numbers are 8*sqrt(3). Note that there was no requirement that the two numbers be distinct. I hope that helps :)

this actually more of a common sense problem. let's say that we change the question to 144 instead then we ask is 144+1 > 12 + 12. Yes! so we know that numbers that are close to eachother have a sum that is smaller. so.... 192 = 2^6 * 3 = 3*2^2 * 2^4 = 16 * 12 this is ofcorse assuming that we are only doing integers. If not then sqrt(192) * sqrt(192) would be the smallest sum