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- Thread starter MarkFL
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Let's begin with the well-known formula for the area of a trapezoid:

\(\displaystyle A=\frac{h}{2}(B+b)\)

We can see that $b=2\text{ ft}$, but $B$ and $h$ are functions of $\theta$.

Using the definition of the sine of an angle \(\displaystyle \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\), we may write:

\(\displaystyle \sin(\theta)=\frac{h}{1}=h\)

Using the definition of the cosine of an angle \(\displaystyle \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\), we may write:

\(\displaystyle B=2+2\cos(\theta)\)

and so the area of the trapezoid as a function of $\theta$ is:

\(\displaystyle A(\theta)=\frac{\sin(\theta)}{2}\left(2+2\cos( \theta)+2 \right)=\sin(\theta)(\cos(\theta)+2)\)

Graphing this function on \(\displaystyle 0\le\theta\le\frac{\pi}{2}\) we see:

and it appears that the area is maximized for approximately \(\displaystyle \theta\approx1.2\)

Using differential calculus, we find:

\(\displaystyle A'(\theta)=\sin(\theta)(-\sin(\theta))+\cos(\theta)(\cos(\theta)+2)=\cos^2(\theta)-\sin^2(\theta)+2\cos(\theta)=2\cos^2(\theta)+2\cos(\theta)-1=0\)

Since we require \(\displaystyle -1\le\cos(\theta)\le1\) the only valid root is:

\(\displaystyle \cos(\theta)=\frac{\sqrt{3}-1}{2}\)

Thus \(\displaystyle \theta=\cos^{-1}\left(\frac{\sqrt{3}-1}{2} \right)\approx1.1960618940861567\)

We really don't need to know the value of $\theta$ though, I just included it for verification of the value the graph indicates.

Since $\theta$ is in the first quadrant, we have:

\(\displaystyle \sin(\theta)=\sqrt{1-\cos^2(\theta)}=\sqrt{1-\left(\frac{\sqrt{3}-1}{2} \right)^2}=\frac{\sqrt{2\sqrt{3}}}{2}=\frac{\sqrt[4]{3}}{\sqrt{2}}\)

and so the maximum area is:

\(\displaystyle A_{\max}=\frac{\sqrt[4]{3}}{\sqrt{2}}\left(\frac{\sqrt{3}-1}{2}+2 \right)=\frac{\sqrt[4]{3}}{2\sqrt{2}}(3+\sqrt{3})\approx2.201834737520805\text{ ft}^2\)