A manufacturer makes a flexible square play yard (Figure 55a), that can be opened at one corner and attached at right angles to a wall or the side of a house, as shown in Figure 55(b). If each side is $3\ {\rm{m}}$ in length, the open configuration doubles the available area from $9\ {\rm{m}}^{2}$ to $18\ {\rm{m}} ^{2}$. Is there a configuration that will more than double the play area?

Figure 55

Solution

Step 1 We want to maximize the play area $A.$

Step 2 Since the play yard must be attached at right angles to the wall, the possible configurations depend on the amount of wall used as a fifth side, as shown in Figure 56. We use $x$ to represent half the length (in meters) of the wall used as the fifth side.

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Figure 56

Step 3 We partition the play area into two sections: a rectangle with area $3(2x) =6x$ and a triangle with base $2x$ and altitude $\sqrt{3^{2}-x^{2}}=\sqrt{9-x^{2}}$. The play area $A$ is the sum of the areas of the two sections. The area to be maximized is $$A=A(x)=6x+\dfrac{1}{2}( 2x)\sqrt{9-x^{2}} =6x+x\sqrt{9-x^{2}}$$

The domain of $A$ is the closed interval $[ 0,3].$

Step 4 To find the maximum area of the play yard, we find the critical numbers of $A$. $$\begin{eqnarray*} {A}' (x)&=&6+x\left[ \dfrac{1}{2}( -2x) ( 9-x^{2}) ^{-1/2}\right] +\sqrt{9-x^{2}}\\[5pt] &=&6-\dfrac{x^{2}}{\sqrt{9-x^{2}}}+\sqrt{9-x^{2}} =\frac{6\sqrt{9-x^{2}}-2x^{2}+9}{\sqrt{9-x^{2}}} \end{eqnarray*}$$

Critical numbers occur when $A^\prime (x) =0$ or where $A^\prime (x)$ does not exist. $A^\prime$ does not exist at $3$, and $A^\prime (x) =0$ at $$\begin{eqnarray*} 6\sqrt{9-x^{2}}-2x^{2}+9 &=&0 \\[3pt] \sqrt{9-x^{2}} &=&\dfrac{2x^{2}-9}{6} \\[3pt] 9-x^{2} &=&\left( \dfrac{2x^{2}-9}{6}\right) ^{\!\!2}=\dfrac{4x^{4}-36x^{2}+81}{ 36} \\[3pt] 324-36x^{2} &=&4x^{4}-36x^{2}+81 \\[3pt] 324 &=&4x^{4}+81 \\[3pt] x^{4} &=&\dfrac{324-81}{4}=\dfrac{243}{4} \\[3pt] x &=&\sqrt[{4}]{{\frac{243}{4}}}\approx 2.792 \end{eqnarray*}$$

Figure 57 Area $\approx$ $19.817 m^2$

The only critical number in the open interval $(0,3)$ is $\sqrt[{ 4}]{{ \dfrac{243}{4}}}\approx 2.792$.

Now we evaluate $A(x)$ at the endpoints $0$ and $3$ and at the critical number $x\approx 2.792$. $$A(0)=0\qquad A(3)=18\qquad A({2.792})\approx 19.817$$

Using a wall of length $2x\approx 2( 2.792) =5.584\ {\rm{m}}$ will maximize the area; the configuration shown in Figure 57 increases the play area by about $10\%$ (from $18$ to $19.817\ {\rm{m}}^{2}$).