A) Find the dimensions of the rectangle with area 324 square inches that has minimum perimeter, and then find the minimum perimeter.

Question

L=______, W=18, minimum perimeter= 72

B) Find the dimensions of the rectangle with perimeter 64 inches that has maximum area, and then find the maximum area

L=_____, W=_____, maximum area=_____

Answer 1

You already have the first part done correctly.

For the second part

L + W = 32

A= L(32-L)

dA/dL = 32 - L

Max area when L = 32 and W = 32 and A = 1024

Answer 2

B)

Approach: You're trying to find the maximum area using one of the rectangle dimensions (length or width) as a variable. Set area as a function of one dimension. Then differentiate your function and set it equal to zero to find your maximum.

You know that your perimeter is 64. Perimeter is twice the length and twice the width:

P = 2L + 2w = 64

And area:

A = L*w

Replace L with a function of w: L = 32-w

Area as a function of w is:

A(w) = w(32-w) = 32w - w²

This function is zero when w = 0 and when w = 32 and has positive values in between. (Graph the function or try some values of w to confirm this), so the maximum is some where between 0 and 32.

Now differentiate your function:

dA/dw = 32 - 2w

Set this equal to zero because at your maximum, the slope of your curve (your A function) is zero.

dA/dw = 32 - 2w = 0

Solve for w:

w = 16

Solve for L:

L = 32-w = 32-16 = 16

Both dimensions are 16. Your rectangle is a square. This means that squares have the most area of any rectangle with the same perimeter.