Need help with Lagrange multipliers?

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

2 Answers

  • 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 derivatives

    dL/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^4

    λ=ye^(xy)/ 5x^4

    From(2)

    xe^(xy)= 5λy^4

    λ=xe^(xy)/ 5y^4 then

    ye^(xy)/ 5x^4=xe^(xy)/ 5y^4 this reduces to

    y/x^4= x/y^4 or

    x^5=y^5

    this has 5 solutions, the only valid solutions is y=x

    From(3)

    64 -x^5-x^5=0

    64-2x^5=0

    64=2x^5

    32=x^5

    x=2 then y=2

    f(2,2)=e^(2*2)=e^4 max

  • Calling the constraint equation g, Lagrange Multipliers ∇f = λ∇g yields

    <ye^(xy), xe^(xy)> = λ<5x^4, 5y^4>.

    Equating like entries yields

    ye^(xy) = 5λx^4

    xe^(xy) = 5λy^4

    If 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!

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