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

In summary, the minimum amount of material needed for the silo to store 1000 ft^3 of hay is approximately 7.816 ft and the dimensions of the silo are a radius of approximately 7.816 ft and a height of 0 ft. This can be achieved by using the formulas for volume and surface area of a cylinder topped with a hemisphere, and solving for the critical value of the surface area. Alternatively, we can use optimization with constraint and Lagrange multipliers to find the same result. However, the fact that a spherical shape encloses more volume per surface area results in the silo having a height of 0 ft and being simply a dome.
  • #1
MarkFL
Gold Member
MHB
13,288
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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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  • #2
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:

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

\(\displaystyle h=0\text{ ft}\)
 

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

1. How do you calculate the surface area of a silo?

The surface area of a silo can be calculated by finding the area of each shape that makes up the silo and then adding them together. For a cylindrical silo, the formula is 2πrh + 2πr^2, where r is the radius and h is the height. For a conical silo, the formula is πrl + πr^2, where r is the radius, l is the slant height, and h is the height.

2. Why is minimizing the surface area of a silo important?

Minimizing the surface area of a silo is important for several reasons. It can help reduce material costs, improve structural stability, and decrease the amount of maintenance and repairs needed. Additionally, a smaller surface area means less exposure to the elements, which can extend the lifespan of the silo.

3. What factors should be considered when trying to minimize the surface area of a silo?

When trying to minimize the surface area of a silo, factors such as the shape, size, and material of the silo should be considered. The type of material being stored in the silo, as well as any environmental factors, should also be taken into account. Other factors such as cost, structural integrity, and ease of construction should also be considered.

4. What are some methods for minimizing the surface area of a silo?

There are several methods for minimizing the surface area of a silo, including using a more efficient shape, such as a dome or square, using thinner materials, and incorporating internal supports or reinforcing features. Other methods may include using specialized coatings or insulation to reduce the need for additional layers or thickness in the silo.

5. Can computer modeling be used to optimize the surface area of a silo?

Yes, computer modeling can be a useful tool for optimizing the surface area of a silo. By inputting various parameters and running simulations, engineers can determine the most efficient shape, size, and materials for the silo. This can also help identify any potential weaknesses or areas that may need additional reinforcement before construction begins.

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