Shape a piece of wire of length and inches into a rectangular shape to maximize the enclosed area. What is the optimum length and width of the rectangle?
Maximize:Area
Restriction:length of the perimeter of the rectangle is equal to L inches.
The most important thing to do first is to get the breadth in terms of the width - or vice versa
Remember, length is a constant and it refers to the perimeter. Since it is a constant I am going to let the length equal 2k.
W = width
B=breadth
W+B=k
W=k-B remember that k is a constant but W and B are variables.
A=WB
A=B(k-B)
A=kB-B^2
Now the idea is to maximize the Area
\(A=kB-B^2\\ \frac{dA}{dB}=k-2B\\ \text{Max or min (stationary points) is when } \frac{dA}{dB} =0\\ but\\ \frac{d^2A}{dB^2}=-2<0\qquad \text{so any stat point is a maximum}\\ \text{Find stat points}\\ \frac{dA}{dB}=0\\ k-2B=0\\ k=2B\\ B=\frac{k}{2}\\ W+B=k\\ W+\frac{k}{2}=k\\ W=\frac{k}{2}\\ \text{The area will be maximum when width = breadth = a quarter of the perimeter.}\\ Max\;area\;\;=k*\frac{k}{2}-\frac{k^2}{4}=\frac{k^2}{4}=\frac{(2k)^2}{16}=\\ Max\;area\;\;=\text{The length of the perimeter squared divided by 16}\\ \)
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