David H. answered 01/22/17
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Patient & Effective Math/Physics Tutor near Fort Worth
So what do we know. Let L be length and W be width.
Perimeter of a rectangle
P = 2L+2W = 60
Thus L = 30-W
The area of a rectangle is
A = LW
Plug in for L
A = (30-W)W = 30W-W2
Take derivative wrt W and set to 0.
dA/dW = 30-2W = 0
Thus W = 15
Take second derivative to see if max or min
d2A/dW2= -2
So our critical point is max sisince second derivative is negative.
So we know L = 30-W=15.
So our dimensions are 15x15 with max area being 225. In fact a square
Graph A = 30W-W2 and you should see this.
Hope helpful. Best - David
Mark M.
01/22/17