Answer
a. $w=6in$, $d=6\sqrt 3\approx 10.4in$,
b. See graph and explanations.
c. See graph and explanations.
Work Step by Step
a. Step 1. We can model the strength with an equation $S=kwd^3, 0\leq w\leq 12,0\leq d\leq 12$, where $k\gt0$ is a constant, $w$ is the width, and $d$ is the depth.
Step 2. Using the Pythagorean Theorem, we have $w^2+d^2=12^2$ or $d=\sqrt {144-w^2}$; thus $S=kw(144-w^2)^{3/2}$,
Step 3. $S'=k(144-w^2)^{3/2}+\frac{3}{2}kw(144-w^2)^{1/2}(-2w)=k(144-w^2)^{1/2}(144-4w^2)$,
Step 4. $S''=\frac{1}{2}k(144-w^2)^{-1/2}(-2w)(144-4w^2)+k(144-w^2)^{1/2}(-8w)=k(144-w^2)^{-1/2}(-144w+4w^3-8\times144w+8w^3)=12kw(144-w^2)^{-1/2}(w^2-108)$.
Step 5. Setting $S'=0$, we get $w=6in$ (discard $w=12$), $d=\sqrt {144-36}=6\sqrt 3\approx 10.4in$, and $S=kwd^3=6(6\sqrt 3)^3k=3888\sqrt 3k\approx6734k$. We can confirm this is a maximum by checking $S''(6)\lt0$, indicating a concave down region with a maximum.
b. With $k=1$, we have $S(w)=w(144-w^2)^{3/2}$ and we can graph the function as shown in the figure. We can find a maximum at $(6,6734)$, which agrees with the results from part (a).
c. Rewriting the function in terms of $d$, we have $S(d)=wd^3=d^3\sqrt {144-d^2}$ and we can graph the function as shown in the figure. We can find a maximum at $(10.4,6734)$ which agrees with the results from part (a).
