Way back in "Calculus of a Single Variable" you learned that max and min of a function may occur where the derivative is 0.
The same is true if you think of the "derivative" as being the gradient: .
Of course, that is the same, as dwsmith says, as requiring that the partial derivatives be equal to 0.
But you also learned that, for a function on a closed, bounded interval, the max or min may occur at the end points (the boundary of the interval) even though the derivative is not 0. Here, the boundary is the square, the four line segments, x= -1, , x= 1, , y= -2, [math\-1\le x\le 1[/tex], and y= 3, .
For example, on the line x= -1, T(-1,y)= 100-4((-1)^2)-(y^2)+((-1)^2)(y^2)= 100- 4 y^2+ y^2= 96. On the line y= 1, T(1, y)= 100- 4(1)^2 -y^2+ (1)^2y^2= 96. On the line y= -2, T(x, -2)= 100- 4x^2- (-2)^2+ x^2(-2)^2= 96, and on the line y= 3, T(x, 3)= 100- 4x^2- (3)^2+ x^2(3)^2= 9+ 5x^2. Look for points on the intervals where the derivatives are 0.
Of course, a max or min might occur at the boundaries of those lines, the vertices, (-1, -2), (1, -2), (1, 3), and (-1, 3).
In general the max or min of a function defined on a two dimensional region may occur at any of three kinds of places:
1) In the interior where the partial derivatives are both 0.
2) On the boundary where the derivative with, respect to whatever parameter you are using, is 0.
3) At the points where two smooth sections of the boundary meet. (At a "corner")
The only way to be sure which points are max or min and what are not is to find all such points and actually evaluate the function at each of them.