Need help with Lagrange multipliers?

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Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If a value does not exist, enter NONE.)f(x, y) = e^(xy); x^5 + y^5 = 64

Mr. Bertha Ziemann IV · Accepted Answer

You construct the Lagrangian L=e^(xy) + λ(64-x^5-y^5)Derive L partially with respect to x, y and λ.The following are partial derivativesdL/dx=ye^(xy) – 5λx^4=0 ..(1)dL/dy=xe^(xy) – 5λy^4=0 ..(2) dL/dλ= 64- x^5 -y^5=0 ..(3)From (1)ye^(xy)= 5λx^4λ=ye^(xy)/ 5x^4From(2) xe^(xy)= 5λy^4λ=xe^(xy)/ 5y^4 thenye^(xy)/ 5x^4=xe^(xy)/ 5y^4 this reduces toy/x^4= x/y^4 orx^5=y^5this has 5 solutions, the only valid solutions is y=xFrom(3)64 -x^5-x^5=064-2x^5=064=2x^532=x^5x=2 then y=2 f(2,2)=e^(2*2)=e^4 max

Mr. Bertha Ziemann IV · Answer

Calling the constraint equation g, Lagrange Multipliers ∇f = λ∇g yields = λ.Equating like entries yieldsye^(xy) = 5λx^4xe^(xy) = 5λy^4If x = 0, then y = 0 (or vice versa).However, we discard this point, because it does not satisfy g(x,y) = x^5 + y^5 = 64.So, we have y/x^4 = 5λe^(-xy) = x/y^4.==> x^5 = y^5==> x = y (only real solution).Substituting this into g yields x^5 + x^5 = 64 ==> x = 2.Hence, we have only one critical point (x, y) = (2, 2).Then, f(2, 2) = e^4 is the maximum, because (64^(1/5), 0) also satisfies g, but f(64^(1/5), 0) = e^0 = 1 < e^4.I hope this helps!

Kale Eichmann

Need help with Lagrange multipliers?

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