Answer
a. See explanations.
b. $x=\frac{51}{8}in$.
c. $L=\frac{51\sqrt 3}{8}\approx11.04in$
Work Step by Step
a. Use the figure given in the Exercise and assign angles $\theta$ and $\phi$ as shown. We have $\phi=\pi-2\theta$. We know the side length $AB=AP+PB=x+x\ cos\phi=8.5$. With $cos\phi=cos(\pi-2\theta)=-cos2\theta=-2cos^2\theta+1=1-2(\frac{x}{L})^2=1-\frac{2x^2}{L^2}$, we have $x+x\ (1-\frac{2x^2}{L^2})=8.5$ which gives $L^2=\frac{2x^3}{2x-8.5}$
b. To minimize $L^2$, let $y=L^2=\frac{2x^3}{2x-8.5}$, we have $y'=\frac{6x^2(2x-8.5)-4x^3}{(2x-8.5)^2}=\frac{8x^3-51x^2}{(2x-8.5)^2}$. Let $y'=0$; discard the solution $x=0$; we get $x=\frac{51}{8}in$. To make sure this $x$-value gives a minimum $L^2$, test signs of $y'$ across the critical point to get $..(-)..(\frac{51}{8})..(+)..$ indicating a concave up region with a minimum.
c. At $x=\frac{51}{8}$, we have $L^2=\frac{2(\frac{51}{8})^3}{2(\frac{51}{8})-8.5}=\frac{3\cdot 51^2}{64}$ and $L=\frac{51\sqrt 3}{8}\approx11.04in$
