Hello Collin,
Let's let $0<W$ be the width (horizontal dimension) of the beam's cross-section and $0<D$ be the depth (vertical dimension). The strength $S$ of the beam is our objective function:
$$S(D,W)=kD^2W$$ where $$0<k\in\mathbb{R}$$ is the constant of proportionality.
The vertices of the beam's cross-section must obviously confined to the circle:
$$x^2+y^2=16^2$$
Hence:
$$W=2x\implies x=\frac{W}{2}$$
$$D=2y\implies y=\frac{D}{2}$$
And so we may write:
$$\left(\frac{W}{2} \right)^2+\left(\frac{D}{2} \right)^2-16^2=0$$
Thus, we may express the constraint as:
$$g(D,W)=D^2+W^2-32^2=0$$
Solving the constraint for $D^2$, we obtain:
$$D^2=32^2-W^2$$
Substituting into the objective function, we find:
$$S(W)=k\left(32^2-W^2 \right)W=k\left(32^2W-W^3 \right)$$
Differentiating with respect to $W$ and equating the result to zero, we find:
$$S'(W)=k\left(32^2-3W^2 \right)=0$$
Taking the positive root, we obtain the critical value:
$$W=\frac{32}{\sqrt{3}}$$
Using the second derivative test to determine the nature of the extremum associated with this critical value, we find:
$$S''(W)=-6kW$$
Since $$0<W$$ then $$S''(W)<0$$ demonstrating that we have found a relative maximum of the objective function. Hence, the dimensions which maximize the beam's strength are:
$$D\left(\frac{32}{\sqrt{3}} \right)=\sqrt{32^2-\left(\frac{32}{\sqrt{3}} \right)^2}=32\sqrt{\frac{2}{3}}$$
$$W=\frac{32}{\sqrt{3}}$$
We could also use a multi-variable technique, Lagrange multipliers.
We have the objective function:
$$S(D,W)=kD^2W$$
subject to the constraint:
$$g(D,W)=D^2+W^2-32^2=0$$
giving rise to the system:
$$2kDW=\lambda(2D)$$
$$kD^2=\lambda(2W)$$
Thus:
$$\lambda=kW=\frac{kD^2}{2W}\implies D^2=2W^2$$
substituting into the constraint, we obtain:
$$2W^2+W^2-32^2=0$$
$$W=\frac{32}{\sqrt{3}}$$
$$D=\sqrt{2}W=32\sqrt{\frac{2}{3}}$$
