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Bari's question at Yahoo! Answers regarding minimizing the surface area of a silo
- Thread starter MarkFL
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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:
\(\displaystyle r=\left(\frac{3\cdot1000\text{ ft}^3}{2\pi} \right)^{\frac{1}{3}}=\sqrt[3]{\frac{1500}{\pi}}\text{ ft}\approx7.816\text{ ft}\)
\(\displaystyle h=0\text{ ft}\)
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:
\(\displaystyle r=\left(\frac{3\cdot1000\text{ ft}^3}{2\pi} \right)^{\frac{1}{3}}=\sqrt[3]{\frac{1500}{\pi}}\text{ ft}\approx7.816\text{ ft}\)
\(\displaystyle h=0\text{ ft}\)