Let w = the width of the rectangle

Let h = the height of the rectangle

Part B:

The area A is to be maximized.

C. A = hw

Use the fact that the perimeter is 60, the amount of fencing.

2h + 2w = 60 –> h = 30 – w

A = (30 – w)w = 30w – w^2

D. dA/dw = 30 – 2w

Set 30 – 2w = 0, w = 15.

E. The critical points are w = 0, w = 30, and w = 15, which we just found.

Clearly w = 0 or w = 30 give zero area (a minimum) and w = 15 gives h = 15 and

an area of A = 15*15 = 225.

Area=B x (30-B)=30B-B^2

For area to be maximum the differential of this has to be zero.So, 30-2B=0: B=15

A square with side = 15 units has the largest area for a perimeter of 60 units.