Answer
$\sqrt {2197}\approx46.9in$
Work Step by Step
Step 1. Use the figure in the Exercise and label the distances as shown, where $x\gt0$.
Step 2. Use the Pythagorean Theorem, $AB^2=(x+27)^2+h^2$, where $y=AB^2$ is the dimension we need to minimize.
Step 3. Using similar triangles of $\Delta ADE$ and $\Delta ABC$, we have $\frac{x}{x+27}=\frac{8}{h}$, leading to $h=\frac{8(x+27)}{x}=8+\frac{216}{x}$
Step 4. Combining the above relations, we have $y=(x+27)^2+(8+\frac{216}{x})^2$. Take the derivative to get $y'=2(x+27)+2(\frac{8(x+27)}{x})(-\frac{216}{x^2})$. Let $y'=0$, we have $(x+27)(1-\frac{8\times 216}{x^3})=0$ which leads to an answers of $x^3=8\times216, x=12in$
Step 5. The shortest dimension is $AB=\sqrt {(x+27)^2+(8+\frac{216}{x})^2}=\sqrt {39^2+26^2}=\sqrt {2197}\approx46.9in$
Step 6.We can confirm the result is a minimum by checking the sign changes of $y'$ across $x=12$ to get $..(-)..(12)..(+)..$
