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!