Question 552093
Recall that the max/min of a quadratic f(x) = ax^2 + bx = c occurs at x = -b/2a.


Dimensions can be assumed to be L and 175-L, since the perimeter is fixed at 350. We want to maximize -L^2 + 175L, in which the maximum occurs at L = 175/2 = 87.5. This makes the rectangle a square.

-------------

Another solution that takes a far different approach. By AM-GM inequality,


*[tex \LARGE \frac{L + (175-L)}{2} \ge \sqrt{L(175-L)}]


This essentially says that the area is at most some constant. Equality occurs if and only if L = 175-L, or L = 87.5 cm.