Question 912065
Going by the above poster's solution:


Area of rectangle must be 300 (500 is impossible -- it can be proven that the maximum area of a rectangle with fixed perimeter is a square).


We want to find L, W such that L+W = 40 and LW = 300. If you notice that 30, 10 work, then you are done. However it could be that L, W are not integers (in a general case), and the easiest solution is to substitute W = 40-L, substitute that into LW = 300 to obtain L(40-L) = 300. You are left with a quadratic equation which you can solve for L.