Question

Determine the dimensions for enclosing the maximum area of a rectangle if:

a. The perimeter is held constant at 280 meters. Assume that the length is greater than or equal to the width.

The length is_______ m.

The width is ________m.


b. The perimeter is held constant at P meters.

The length is_______ m.


The width is_________ m.

Answer 1

Let L be the length and W be the width. The area A is:\[A=L·W\]and the perimeter P is:\[P=2(W+L)\rightarrow L=\frac{ P }{ 2 }-W\]Then we have:\[A=\left( \frac{ P }{ 2 } -W\right)W=\frac{ P }{ 2 }W-W^2\]In order to enclose the maximum area we need to solve:\[\frac{ dA }{ dW }=\frac{ P }{ 2 }-2W=0\rightarrow W=\frac{ P }{ 4}\rightarrow L=\frac{ P }{ 2}-\frac{ P}{ 4 }=\frac{ P}{4 }\]
As\[L=W=\frac{ P }{ 4 }\], it is a square

Answer 2

for P=280 we get L=W=280/4=70

Answer 3

Thank you so much you were right:)

Answer 4

welcome