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
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)
λ=xe^(xy)/ 5y^4 then
ye^(xy)/ 5x^4=xe^(xy)/ 5y^4 this reduces to
y/x^4= x/y^4 or
this has 5 solutions, the only valid solutions is y=x
x=2 then y=2
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!