Question

Finding the sides of a rectangle with only a known perimeter

Answer 1

The perimeter of a rectangle is 284ft. What is the length and width of the rectangle and what is the maximum area?

Answer 2

`The perimeter of a rectangle is 284ft`

so

P = 2L + 2W
P = 2*(L+W)
284 = 2*(L+W)
142 = L+W
L = -W+142

now turn to the area of a rectangle formula

A = L*W

and plug in L = -W+142 to get

A = (-W+142)*W
A = -W^2 + 142W

if you treat W as x and A as y, then we have

y = -x^2 + 142x

the goal is to find the vertex of this parabola

Answer 3

So completing the square?

Answer 4

use x = -b/(2a) and tell me what you get

Answer 5

correct

Answer 6

I got -142

Answer 7

y = -x^2 + 142x is the same as y = -1x^2 + 142x

a = -1
b = 142

what is the value of x = -b/(2a) ?

Answer 8

x = -b/(2a)
x = -(142)/2(-1) = -142/-2 = 71

Answer 9

now use this to find y

plug x = 71 into y = -x^2 + 142x

Answer 10

I got 5041. It looks very big

Answer 11

the value x = 71 corresponds to W (since we replaced W with x), so the width is W = 71

the length is

L = -W + 142
L = -71 + 142
L = 71

the width and length are equal so this rectangle is also a square

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when you got 5041, that is the area of the rectangle with side lengths of 71 and 71

notice how 71*71 = 71^2 = 5041

Answer 12

Oh, okay, I see. So L and W are both 71 and the max area is 5041?

Answer 13

it turns out that if you have a fixed perimeter, and you want to max out a rectangle area, then that rectangle will turn into a square

Answer 14

correct

Answer 15

so a quick shortcut is to compute L = P/4

in this case, P = 284

Answer 16

Thanks so much for your help, I really appreciate it :)

Answer 17

no problem