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

Answer 2

P = 2x + 2y, A = xy

=>P = 2x + 2A/x

dP/dx = 0 for max. find value of x where that occurs.

Answer 3

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

Answer 4

@heavenfun13 you should read the code of conduct and help students do the work themselves, not do it for them

Answer 5

i did