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Bari's question at Yahoo! Answers regarding minimizing the surface area of a silo

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MarkFL

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Feb 24, 2012
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Here is the question:

Can you solve this Calculus problem?


I need help.
Solve using Related Rates or Optimization.

Bob is building a silo to store 1000 ft^3 of hay. Find the minimum amount of material and the dimensions of the silo to achieve it. Since his silo is a cylinder topped with a hemisphere, the formulas are as follows:
V=pi x r^2 x h + (2/3)pi x r^3
SA= 2 x pi x r x h + 2 x pi x r ^2
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 Bari,

The volume $V$ of the silo is given by:

\(\displaystyle V=\pi hr^2+\frac{2}{3}\pi r^3\)

The surface area $S$ is given by:

\(\displaystyle S=2\pi hr+2\pi r^2\)

Solving the first equation for $h$, we obtain:

\(\displaystyle h=\frac{3V-2\pi r^3}{3\pi r^2}\)

Substituting for $h$ into the second equation, we obtain the surface area as a function of one variable, $r$:

\(\displaystyle S(r)=2Vr^{-1}+\frac{2\pi}{3}r^2\)

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

\(\displaystyle S'(r)=-2Vr^{-2}+\frac{4\pi}{3}r=0\)

Solving for $r$, we obtain the critical value:

\(\displaystyle r=\left(\frac{3V}{2\pi} \right)^{\frac{1}{3}}\)

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

\(\displaystyle S''(r)=4Vr^{-3}+\frac{4\pi}{3}>0\,\forall\,0<r\)

Hence, the extremum is a minimum. And so the dimensions if the silo which minimizes the surface area are:

\(\displaystyle r=\left(\frac{3V}{2\pi} \right)^{\frac{1}{3}}\)

\(\displaystyle h=\frac{3V-2\pi\left(\frac{3V}{2\pi} \right)}{3\pi\left(\frac{3V}{2\pi} \right)^{\frac{2}{3}}}=0\)

Another method we could use is optimization with constraint, via Lagrange multipliers.

We have the objective function:

\(\displaystyle S(h,r)=2\pi hr+2\pi r^2\)

Subject to the constraint:

\(\displaystyle g(h,r)=\pi hr^2+\frac{2}{3}\pi r^3-V=0\)

which yields the system:

\(\displaystyle 2\pi r=\lambda\left(\pi r^2 \right)\implies \lambda=\frac{2}{r}\)

\(\displaystyle 2\pi h+4\pi r=\lambda\left(2\pi hr+2\pi r^2 \right)\implies \lambda=\frac{h+2r}{r(h+r)}\)

This implies:

\(\displaystyle \frac{2}{r}=\frac{h+2r}{r(h+r)}\)

\(\displaystyle 2hr+2r^2=hr+2r^2\)

\(\displaystyle hr=0\)

We know $r\ne0$, hence we must have $h=0$, and so the constraint becomes:

\(\displaystyle \frac{2}{3}\pi r^3-V=0\implies r=\left(\frac{3V}{2\pi} \right)^{\frac{1}{3}}\)

Note: The fact the $h=0$ follows from the fact that a spherical shape encloses more volume per surface area than a cylinder, and so the silo winds up simply being a dome.

Now, using the given data:

\(\displaystyle V=1000\text{ ft}^3\)

the dimensions we seek are:

\(\displaystyle r=\left(\frac{3\cdot1000\text{ ft}^3}{2\pi} \right)^{\frac{1}{3}}=\sqrt[3]{\frac{1500}{\pi}}\text{ ft}\approx7.816\text{ ft}\)

\(\displaystyle h=0\text{ ft}\)