Tom K. answered • 07/26/20
Knowledgeable and Friendly Math and Statistics Tutor
Ler's consider the boundaries and interior.
Clearly, the minimum is 0 at (0,0) as x^2+y^2 >= y^2 >= y^2x^3 on [0,1]x[0,1], so x^2+y^2 - y^2x^3 >= 0, and it equals 0 at (0,0).
If x = 0 or y = 0, x^2 + y^2 <= 1, so the maximum value on x = 0 or y = 0 is 1 (at 0, 1) and (1, 0)
If x = 1, we have 1 + y^2 - y^2 = 1.
If y = 1, we have 1 + x^2 - x^3.
The derivative with respect to x is 2x - 3x^2 = 3x(x - 2/3). We have already evaluated x = 0. Also, note that the second derivative = 2 - 6x = -2 at x = 2/3 means a maximum.
When x = 2/3 and y = 1, x^2 + y^2 - y^2x^3 = 1 + 4/9 - 8/27 = 1 4/27
Now, let's consider the interior.
∂f(x,y)/∂x = 2x - 3x^2y^2
∂f(x,y)/∂y = 2y - 2yx^3 = 2y(1-x^3)
∂f(x,y)/∂y = 0 at y = 0 (already examined) and x=1(both on the boundary). Thus, we don't need to look at the other partial because it is irrelevant.
Thus, our minimum is 0 at (0,0) and our maximum is 1 4/27 or 31/27 at (2/3,1).
