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Among all the rectangles whose perimeters are 100 ft, find the dimensions of the one with the maximum area.

AdamTaurus  Oct 24, 2017
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7+0 Answers

 #1
avatar+86649 
+2

I'm a little tired......but the area will be maximized when the rectangle is a square with a side of 25

 

I'll go through it, AT, if you want me to........

 

 

cool cool cool

CPhill  Oct 24, 2017
 #2
avatar+771 
+1

It's fine. You don't have to if you don't feel like it. I'm sure someone else on here can explain it.

AdamTaurus  Oct 24, 2017
 #3
avatar+7056 
+1

Let's call the length of the rectangle  L, and the width  W .

 

perimeter  =  L + L + W + W

 

perimeter  =  2L  +  2W

 

100  =  2L  +  2W            Solve this for  L .

 

100 - 2w  =  2L        Divide through by  2 .

 

50 - W  =  L

 

 

area  =  L * W           Substitute  50 - W  in for  L .

 

area  =  (50 - W) * W

 

area  =  50W - W2

 

area  =  -W2 + 50W

 

The maximum area will be when  W = - 50 / [2(-1)]  =  25

 

and L  =  50 - W  =  50 - 25  =  25

hectictar  Oct 24, 2017
 #4
avatar+771 
+2

Thanks Hectictar!

laugh

AdamTaurus  Oct 24, 2017
 #5
avatar+86649 
+2

Here it is

 

The perimeter is given by

 

P  =  2 ( W + L)

 

P / 2  =   W + L    ⇒   P/2 - W  =  L     (1)

 

And the area is

 

A =  L  * W        sub  (1)  for L

 

A =  (P/2 - W) * W        simplify

 

(P/2) W  - W^2     rearrange

 

-1W^2  + (P/2)W

 

a =  -1    b  =  (P/2)

 

Again....the  "W"  that maximizes  the area is given by     -b / [ 2a]  =

 

- ( P/2) / [2 ( -1) ]  =  - P/ -4  =  P /4

 

And when  W  =  P/4       L  =  P/2 - P/4  =  P/4

 

So.....  W = L   =   P/4  =  100 / 4  =   25  ft

 

This will always be true, AT......for any perimeter......area is maximized  when  W = L  = P/4

 

 

cool cool cool

CPhill  Oct 24, 2017
 #7
avatar+771 
+1

Thanks CPhill! I appreciate it.

AdamTaurus  Oct 24, 2017
 #6
avatar+86649 
+1

Thanks, Hectictar.....!!!!

 

 

cool cool cool

CPhill  Oct 24, 2017

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