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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}\\ $

Shape a piece of wire of length and inches into a rectangular shape to maximize the enclosed area. {nl} What is the optimum length and width of the rectangle? {nl} Maximize:Area {nl} Restriction:length of the perimeter of the rectangle is equal to L inches. {nl} A=lw {nl} P=2l+2w

Let l = length of the rectangle

Let w = width of the rectangle

Let L = perimeter of the rectangle

$\begin{array}{|rcll|} \hline 2\cdot ( l + w ) &=& L \\ l+w &=& \frac{L}{2} \\ w &=& \frac{L}{2} - l \\\\ A &=& l\cdot w \\ A &=& l\cdot \left( \frac{L}{2} - l \right) \\ A &=& l\cdot \frac{L}{2} - l^2 \\\\ A' &=& \frac{L}{2} - 2\cdot l = 0 \\ \frac{L}{2} - 2\cdot l &=& 0 \\ \frac{L}{2} &=& 2\cdot l \\ L &=& 4\cdot l \\ \mathbf{l} & \mathbf{=} & \mathbf{ \frac{L}{4} } \\\\ A'' &=& - 2 \qquad \text{ max.}\\\\ w &=& \frac{L}{2} - l \\ w &=& \frac{L}{2} - \frac{L}{4} \\ \mathbf{w} & \mathbf{=} & \mathbf{ \frac{L}{4} } \\\\ \hline \end{array}$

The optimum length = width = $\frac{L}{4}$

The max. area is  $l \cdot w = \frac{L}{4} \cdot \frac{L}{4} = \frac{L^2}{16}$

Maximize:Area