A rectangle is inscribed in a semicircle of radius $2$. Find the dimensions of the rectangle that has the maximum area.

Solution We present two methods of solution: The first uses analytic geometry, the second uses trigonometry. To begin, we place the semicircle with its diameter along the $x$-axis and center at the origin. Then we inscribe a rectangle in the semicircle, as shown in Figure 59. The length of the inscribed rectangle is $2x$ and its height is $y.$

Figure 59

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Analytic Geometry Method The area $A$ of the inscribed rectangle is $A=2xy$ and the equation of the semicircle is $x^{2}+y^{2}=4$, $y\geq 0.$ We solve for $y$ in $x^{2}+y^{2}=4$ and obtain $y=\sqrt{4-x^{2}}$. Now we substitute this expression for $y$ in the area formula for the rectangle to express $A$ as a function of $x$ alone. $$\begin{eqnarray*} A&=&A(x)= 2x\sqrt{4-x^{2}}\qquad 0\leq x\leq 2\\[-6.4pt] &&\hspace{-1.8pc}\color{#0066A7}{\underset{\hbox{\(A=2xy; y=\sqrt{4-x^{2}}\)}}{\uparrow}} \end{eqnarray*}$$

Then $$\begin{eqnarray*} {A}' (x) &=&2\left[ x\left( {{\dfrac{1}{2}}}\right) \ (4-x^{2}) ^{-1/2}(-2x)+\sqrt{4-x^{2}}\right] =2\left[ \dfrac{-x^{2}}{ \sqrt{4-x^{2}}}+\sqrt{4-x^{2}}\right]\\[3pt] &=&2\left[ \frac{-x^{2}+( {4-x^{2}}) }{\sqrt{4-x^{2}}}\right] =2\left[ \frac{-2({x^{2}-2})}{\sqrt{4-x^{2}}}\right] =-\frac{4({x^{2}-2})}{\sqrt{4-x^{2}}} \end{eqnarray*}$$

The only critical number in the open interval $(0,2)$ is $\sqrt{2}$, where $A^\prime ( \sqrt{2}) =0.$ [$-\sqrt{2}$ and $-2$ are not in the domain of $A$, and $2$ is not in the open interval $( 0,2)$.] The values of $A$ at the endpoints $0$ and $2$ and at the critical number $\sqrt{2}$ are $$A(0)=0\qquad A(\sqrt{2})=4\qquad A(2)=0$$

The maximum area of the inscribed rectangle is $4$, and it corresponds to the rectangle whose length is $2x=2\sqrt{2}$ and whose height is $y=\sqrt{2}$.

Figure 60

Trigonometric Method Using Figure 59, we draw the radius $r=2$ from $O$ to the vertex of the rectangle and place the angle $\theta$ in the standard position, as shown in Figure 60. Then $x=2\cos \theta$ and $y=2\sin \theta ,$ $0\leq \theta \leq \dfrac{\pi }{2}$. The area $A$ of the rectangle is $$\begin{eqnarray*} A&=&2xy=2(2\, \cos \theta )(2\, \sin\, \theta )=8\, \cos \theta\, \sin\, \theta \underset{\underset{{\color{#0066A7}{\hbox{\(\sin (2\theta) =2\, \sin\, \theta \cos \theta\)}}}}{\color{#0066A7}{\uparrow}}}{=} 4\sin ( 2\, \theta )\\[-6.3pt] \end{eqnarray*}$$

Since the area $A$ is a differentiable function of $\theta$, we obtain the critical numbers by finding $A^\prime (\theta )$ and solving the equation $A^\prime (\theta ) =0$. $$\begin{eqnarray*} A^\prime (\theta ) &=&8\, \cos ( 2\theta ) =0 \\[1pt] \cos ( 2\theta ) &=&0 \\[1pt] 2\theta &=&\cos ^{-1}0=\dfrac{\pi }{2} \\[3pt] \theta &=&\dfrac{\pi }{4} \end{eqnarray*}$$

We now find $A^{\prime \prime} (\theta )$ and use the Second Derivative Test. $$\begin{eqnarray*} {A}'' (\theta )&=&-16\, \sin ( 2\theta ) \\[3pt] {A}'' \left( {\dfrac{\pi }{4}}\right) &=&-16\, \sin \left( 2\cdot \dfrac{\pi }{4}\right) =-16\, \sin \dfrac{\pi }{2}=-16<0 \end{eqnarray*}$$

So at $\theta =\dfrac{\pi }{4}$, the area $A$ is maximized and the maximum area of the inscribed rectangle is $$A\left( \dfrac{\pi }{4}\right) =4\sin \left( 2\cdot \frac{\pi }{4}\right) =4 \hbox{ square units}$$

The rectangle with the maximum area has $$\hbox{length}\, 2x=4\cos \dfrac{\pi }{4}=2 \sqrt{2}\,\, \hbox{and height}\,\, y=2\, \sin \dfrac{\pi }{4} =\sqrt{2}$$