Maximizing Area

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 \(y.\)

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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 \(y=\sqrt{2}\).

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 \[ \hbox{length}\, 2x=4\cos \dfrac{\pi }{4}=2 \sqrt{2}\,\, \hbox{and height}\,\, y=2\, \sin \dfrac{\pi }{4} =\sqrt{2} \]