Collin's question at Yahoo Answers regarding maximization of beam strength

[MarkFL]

Calculus Max Min Application Problem?


If the strength of a rectangular beam is proportional to the product of its width and the square of its depth, find the dimensions of the strongest beam that could be cut from a log whose cross section is a circle of the form x^2+y^2=256

Please show work, thanks.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

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:

$$2W^2+W^2-32^2=0$$

$$W=\frac{32}{\sqrt{3}}$$

$$D=\sqrt{2}W=32\sqrt{\frac{2}{3}}$$