Question

A rectangle has an area of 16 square feet. Its length and width are whole numbers. What is its maximum perimeter?

Answer 1

(L) x (W) = 16

2L + 2W = P

Solve for one of the variables, in this case lets solve for (w)

LW = 16

W = 16/L

plug that in to the Perimeter Eqauation:

2(L) + 2( 16/L ) = P

Now take the derivative of P

P = 2L + 32/L

P'= 2 + 32( -1/L^2 )

Now set P' to 0, because the derivative is 0 at its Maximums and Minimums

0 = 2 - 32/(L^2)

-2 = -32/(L^2)

-2(L^2) = -32

L^2 = 16

L = 4

Plug this back in to solve for W

W = 16/L

W = 16/4

W = 4

tada! so the maximum perimeter must be 16, and the dimensions are 4 by 4

Answer 2

4x4, 8x2, 16x1 possibilities

Answer 3

Actually I found the answer to be 34 feet :) But thanks for the help!