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

AI Thread Summary
To minimize the surface area of a silo storing 1000 ft³ of hay, the silo's optimal shape is determined to be a hemisphere with a radius of approximately 7.816 ft and a height of 0 ft. The volume and surface area formulas for the cylindrical and hemispherical components are used to derive these dimensions. By applying calculus techniques, including differentiation and the second derivative test, it is confirmed that the minimum surface area occurs when the height is zero, indicating a dome shape. The analysis also suggests that a spherical shape is more efficient in enclosing volume relative to surface area compared to a cylinder. Thus, the optimal design for the silo is a hemisphere with the specified radius.
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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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Hello Bari,

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

$$V=\pi hr^2+\frac{2}{3}\pi r^3$$

The surface area $S$ is given by:

$$S=2\pi hr+2\pi r^2$$

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

$$h=\frac{3V-2\pi r^3}{3\pi r^2}$$

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

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

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

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

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

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

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

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

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

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

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

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

We have the objective function:

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

Subject to the constraint:

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

which yields the system:

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

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

This implies:

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

$$2hr+2r^2=hr+2r^2$$

$$hr=0$$

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

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

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:

$$V=1000\text{ ft}^3$$

the dimensions we seek are:

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

$$h=0\text{ ft}$$
 
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