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How can I solve this optimization calculus problem?

Find the dimensions of the rectangle of area A square units that has the smallest perimeter.



I think I have to find the area in terms of the perimeter so when I do that, I get:

A = (1/2)(P-2w)w so,

A' = (1/2) [(P-2w) + (dP/dA +2)]

The problem that I run into is how to graph the derivative of the equation since there are two variables, P and w.

Public Comments

1. Recall that area A = lw (length * width),
   
   and perimeter p=2l+2w. You want the minimum perimeter for a given area.
   
   
   
   Since p=2l+2w, w=(p-2l)/2
   
   
   
   A=lw = l(p-2l)/2
   
   A = (1/2)(pl - 2l^2)
   
   A' = (1/2)(p -4l), which is set equal to 0 to give you the minimum.
   
   
   
   (1/2)(p-4l)=0
   
   p-4l=0
   
   p=4l
   
   
   
   So the perimeter is minimized when all sides are the same (or, in other words, a square has the smallest perimeter for the given area).
   
   
   
   Hope that helps.