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

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MarkFL

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Feb 24, 2012
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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.
 
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MarkFL

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Feb 24, 2012
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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:

\(\displaystyle S(D,W)=kD^2W\) where \(\displaystyle 0<k\in\mathbb{R}\) is the constant of proportionality.

The vertices of the beam's cross-section must obviously confined to the circle:

\(\displaystyle x^2+y^2=16^2\)

Hence:

\(\displaystyle W=2x\implies x=\frac{W}{2}\)

\(\displaystyle D=2y\implies y=\frac{D}{2}\)

And so we may write:

\(\displaystyle \left(\frac{W}{2} \right)^2+\left(\frac{D}{2} \right)^2-16^2=0\)

Thus, we may express the constraint as:

\(\displaystyle g(D,W)=D^2+W^2-32^2=0\)

Solving the constraint for $D^2$, we obtain:

\(\displaystyle D^2=32^2-W^2\)

Substituting into the objective function, we find:

\(\displaystyle S(W)=k\left(32^2-W^2 \right)W=k\left(32^2W-W^3 \right)\)

Differentiating with respect to $W$ and equating the result to zero, we find:

\(\displaystyle S'(W)=k\left(32^2-3W^2 \right)=0\)

Taking the positive root, we obtain the critical value:

\(\displaystyle W=\frac{32}{\sqrt{3}}\)

Using the second derivative test to determine the nature of the extremum associated with this critical value, we find:

\(\displaystyle S''(W)=-6kW\)

Since \(\displaystyle 0<W\) then \(\displaystyle S''(W)<0\) demonstrating that we have found a relative maximum of the objective function. Hence, the demensions which maximize the beam's strength are:

\(\displaystyle D\left(\frac{32}{\sqrt{3}} \right)=\sqrt{32^2-\left(\frac{32}{\sqrt{3}} \right)^2}=32\sqrt{\frac{2}{3}}\)

\(\displaystyle W=\frac{32}{\sqrt{3}}\)

We could also use a multi-variable technique, Lagrange multipliers.

We have the objective function:

\(\displaystyle S(D,W)=kD^2W\)

subject to the constraint:

\(\displaystyle g(D,W)=D^2+W^2-32^2=0\)

giving rise to the system:

\(\displaystyle 2kDW=\lambda(2D)\)

\(\displaystyle kD^2=\lambda(2W)\)

Thus:

\(\displaystyle \lambda=kW=\frac{kD^2}{2W}\implies D^2=2W^2\)

substituting into the constraint, we obtain:

\(\displaystyle 2W^2+W^2-32^2=0\)

\(\displaystyle W=\frac{32}{\sqrt{3}}\)

\(\displaystyle D=\sqrt{2}W=32\sqrt{\frac{2}{3}}\)