Question

Find the maximum value of the function f(x,y) = 9x2−9xy+y3 on the closed rectangle whose corners are the points (1,1), (1,−1), (−1,1) and (−1,−1)

Answer 1

f(-1,y) = y^3+9y+9
f(1,y) = y^3-9y+9
f(x,-1) = 9x^2+9x-1
f(x,1) = 9x^2-9x+1
So, check for maxima of those on the interval [-1,1]

Then, you have
∂f/∂x = 18x-9y
∂f/∂y = 3y^2-9x
Extrema occur when both these are zero.
Along the line ∂f/∂x=0 or y=2x,
g(x) = f(x,y) = 8x^3-9x^2
g'(x) = 24x^2-18x = 6x(4x-3)
So, at (3/4,3/2) there is an extremum

You can do the other case, and use the 2nd derivative tests to determine whether it's a max or min.