Wait, wah...? Partial Fractions and Derivative Proof, I think..?

I feel like we need more math questions in here.

In this thread?

I don't have that fancy math-writing program, sorry if it's sloppy

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. < what does this mean? I thought it was there if there were more than one local min/max. If an answer does not exist, enter DNE.)

f(x,y) = xy + 64/x + 64/y

Local Maximum values(s) : DNE [correct]

Local Minimum values(s) : 48 Fixed it; works now

Saddle points(s) (x, y, f) = DNE [correct]

I think I'm putting the answer in the wrong form..?

First derivatives:
fx = y - 64/x^2; fy = x - 64/y^2

Second derivatives:
fxx = 128/x^3; fyy = 128/y^3; fxy = 1

Set first derivatives = 0, substitute:
y = 64/x^2; x = 64/y^2
x = 64/(64/x^2)^2 = x^4/64

Solve for x:
x(x^3 - 64) = 0
x = 0, x = 4

Function is undefined at x = 0.

Substituting x = 4 into fx gives y = 4.

D(x,y) = fxx(x,y)*fyy(x,y) - fxy(x,y)
D(4,4) = (2)(2) - 1 = 3

Since D(4,4) > 0 and fxx(4,4) > 0, f(4,4) is a local minimum. Therefore, there are no local maximum or saddle points.

WebAssign is not accepting (4,4) as an answer. I have tried (4,4,48) [for (x,y,f) format?], f(4,4), f(4,4)=48. Is there a standard form for answering that I missed?


EDIT: 48 worked. Guess value(s) should have given it away, since it wasn't asking for a point.

I'll probably be looking for more math help next semester and post a few questions.