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Let half the width of the rectangle  =  w

Then half the length of the rectangle  =  √[ 1 - w² ]

And let the area of the rectangle  =  A

 

\(A\ =\ 4w\sqrt{1-w^2}\\~\\ \frac{dA}{dw}\ =\ 4\frac{d}{dw}\big(w\sqrt{1-w^2}\,\big)\\~\\ \frac{dA}{dw}\ =\ 4\Big[(w)(\frac12)(1-w^2)^{-\frac12}(-2w)+(\sqrt{1-w^2})(1)\Big]\\~\\ \frac{dA}{dw}\ =\ 4\Big[\frac{-w^2}{\sqrt{1-w^2}}+\sqrt{1-w^2}\Big]\\~\\ \frac{dA}{dw}\ =\ 4\Big[\frac{-w^2}{\sqrt{1-w^2}}+\frac{1-w^2}{\sqrt{1-w^2}}\Big]\\~\\ \frac{dA}{dw}\ =\ \frac{4-8w^2}{\sqrt{1-w^2}}\)

 

Now let's find what value of  w  makes  \(\frac{dA}{dw}\)  be  0 .

 

\(0\ =\ \frac{4-8w^2}{\sqrt{1-w^2}}\\~\\ 0\ =\ 4-8w^2\qquad\text{and}\qquad\sqrt{1-w^2}\ \neq\ 0\qquad\text{that is}\qquad w\ \neq\ \pm1\\~\\ 8w^2\ =\ 4\\~\\ w^2\ =\ \frac12\\~\\ w\ =\ \sqrt{\frac12}\qquad\text{or}\qquad w\ =\ -\sqrt{\frac12} \)

 

We know half the width of the rectangle can't be negative,

and by looking at a graph https://www.desmos.com/calculator/txezockcml

we can confirm that the maximum value of  A  occurs when  \(w=\sqrt{\frac12}\) .

 

When   \(w=\sqrt{\frac12}\)  ,

 

\(A\ =\ 4w\sqrt{1-w^2}\ =\ 4\sqrt{\frac12}\sqrt{1-\frac12}\ =\ 4\sqrt{\frac12}\sqrt{\frac12}\ =\ 4\cdot\frac12\ =\ 2\)

 

The maximum area is  2  sq units.