Answer
a. $w=4\sqrt 3\approx 6.93in$, $d=4\sqrt 6\approx 9.80in$
b. See graph and explanations.
c. See graph and explanations.
Work Step by Step
a. We can model the strength with an equation $S=kwd^2$ where $k\gt0$ is a constant. Using the Pythagorean Theorem, we have $w^2+d^2=12^2$ or $d^2=144-w^2$; thus $S=kw(144-w^2)=k(144w-w^3)$, $S'=k(144-3w^2)$, and $S'=-6kw$. Setting $S'=0$, we get $w=4\sqrt 3\approx 6.93in$, $d=\sqrt {144-\frac{144}{3}}=4\sqrt 6\approx 9.80in$, and $S=kwd^2=4\sqrt 3\times16\times6k=384\sqrt 3\approx665.1k$.
We can confirm that this is a maximum by checking $S''=-6kw=-24\sqrt 3k\lt0$, indicating a concave down region with a maximum.
b. With $k=1$, we have $S=144w-w^3$ and we can graph the function as shown in the figure. We can find a maximum at $(6.93,665.1)$, which agrees with the results from part (a).
c. Rewrite the function in terms of $d$; we have $S=wd^2=d^2\sqrt {144-d^2}$ and we can graph the function as shown in the figure. We can find a maximum at $(9.80,665.1)$, which agrees with the results from part (a).
