Problem 9 - Active Calculus

Question

A company is designing propane tanks that are cylindrical with hemispherical ends. Assume that the company wants tanks that will hold 1000 cubic feet of gas, and that the ends are more expensive to make, costing \(\$ 5\) per square foot, while the cylindrical barrel between the ends costs \(\$ 2\) per square foot. Use calculus to determine the minimum cost to construct such a tank.

Step-by-Step Solution

Answer

The minimum cost is approximately $1714.45.

Step 1 - Define Variables

Let the radius of the cylinder and hemispheres be denoted as \( r \), and the height of the cylindrical part be denoted as \( h \). The volume constraints and cost equations need to incorporate these variables.

Step 2 - Write Volume Equation

The volume of the entire tank is the sum of the volume of the cylindrical part and the volumes of two hemispheres. The volume of the cylinder is \( \pi r^2 h \) and the volume of two hemispheres is \( \frac{4}{3} \pi r^3 \). The total volume is given by: \[ V = \pi r^2 h + \frac{4}{3} \pi r^3 \] We are given that the total volume is 1000 cubic feet, so: \[ \pi r^2 h + \frac{4}{3} \pi r^3 = 1000 \]

Step 3 - Solve for h

Rearrange the volume equation to solve for \( h \): \[ h = \frac{1000 - \frac{4}{3} \pi r^3}{\pi r^2} \]

Step 4 - Surface Area and Cost Equations

Calculate the surface areas. The surface area of the cylindrical part is \(2\pi r h\) and the surface area of two hemispheres (which make a complete sphere) is \(4\pi r^2\). The respective cost for each part is then: Cost of cylindrical part: \[ C_{cyl} = 2\pi r h \times 2 \quad (\text{since the cost is } \$2 \text{ per square foot}) \] Cost of hemispheres: \[ C_{sphere} = 4\pi r^2 \times 5 \quad (\text{since the cost is } \$5 \text{ per square foot}) \]

Step 5 - Total Cost Function

Combine the cost equations to write the total cost function \(C\): \[ C = 4\pi r h + 20\pi r^2 \] Substitute the expression for \( h \) found in Step 3: \[ C = 4\pi r \left( \frac{1000 - \frac{4}{3} \pi r^3}{\pi r^2} \right) + 20\pi r^2 \] Simplify the expression: \[ C = \frac{4000}{r} + \left( \frac{16}{3} \right) \pi^2 r^3 \]

Step 6 - Differentiate Total Cost Function

Find the derivative of \( C \) with respect to \( r \): \[ C' = -\frac{4000}{r^2} + \left( \frac{16}{3} \right) 3\pi^2 r^2 \] Simplify the derivative: \[ C' = -\frac{4000}{r^2} + 16\pi^2 r^2 \]

Step 7 - Critical Points

Set the derivative \( C' \) equal to zero to find critical points: \[ -\frac{4000}{r^2} + 16\pi^2 r^2 = 0 \] Solve for \( r \): \[ 16\pi^2 r^4 = 4000 \] \[ r^4 = \frac{4000}{16\pi^2} \] \[ r = \left( \frac{250}{\pi^2} \right)^{1/4} \approx 3.63 \text{ feet} \]

Step 8 - Determine Corresponding h

Substitute \( r \approx 3.63 \text{ feet} \) back into the equation to find \( h \): \[ h = \frac{1000 - \frac{4}{3} \pi (3.63)^3}{\pi (3.63)^2} \approx 7.25 \text{ feet} \]

Step 9 - Minimum Cost Calculation

Substitute \( r \) and \( h \) back into the cost function to find the minimum cost: \[ C = 4\pi (3.63)(7.25) + 20\pi (3.63)^2 \approx 885.07 + 829.38 \approx 1714.45 \text{ dollars} \]

Key Concepts

Volume and Surface Area Calculation

Understanding the geometry of the propane tank is essential. The tank comprises a cylindrical section and two hemispherical ends. To find the volume and surface area, we start by defining the dimensions of this composite shape. The radius of the cylinder and the hemispheres is denoted by \( r \), and the height of the cylindrical part is \( h \). The volume of the whole tank is the sum of the volumes of the cylinder and two hemispheres.

The volume of the cylindrical part is given by:
\(\[ V_{cyl} = \,\pi \,r^2 \,h \]\)

The volume of the two hemispheres, which together form a sphere, is:
\(\[ V_{sphere} = \frac{4}{3}\pi \,r^3\]\)

Combining these, the total volume of the tank is:
\(\[ V = \,\pi \,r^2 \,h \, +\,\frac{4}{3} \,\pi \,r^3\]\)

Given in the problem, the tank must hold 1000 cubic feet of gas. So, we set up the equation:
\(\[ \,\pi \,r^2 \,h \, +\,\frac{4}{3} \,\pi \,r^3 \,= \,1000\]\)

To solve for the height \( h \), we rearrange the equation:
\(\[ h \,=\,\frac{1000 \, -\,x\frac{4}{3}\pi \,r^3}{\pi \, r^2}\]\)

Regarding surface area, the cylindrical part's surface area (excluding the top and bottom circles) is:
\(\[ A_{cyl} \, =\,2\pi \,rh\]\)

The surface area of two hemispheres, which together form a complete sphere, is:
\(\[ A_{sphere} \, =\,4\pi \,r^2\]\)

With these calculations, we move on to determining the cost.

Cost Function

The cost of constructing different parts of the tank varies. The cylindrical barrel is cheaper to make, costing \(\$2\) per square foot, while the hemispherical ends cost more at \(\$5\) per square foot. To find the total cost, we need to calculate the cost for each part separately and then sum them up.

The cost of constructing the cylindrical part is calculated using its surface area:
\(\[ C_{cyl} \, =\,2\pi \, r \,h \, \times \, 2\]\)

Here, we've multiplied the surface area by \(2\) dollars per square foot.

The cost of the hemispherical ends is obtained from their surface area:
\(\[ C_{sphere} \, =\,4\pi \, r^2 \, \times \, 5\]\)

Here, we multiplied by \(5\) dollars per square foot. Combining these results, the total cost function \( C \) in terms of radius \( r \) and height \( h \) is:

\(\[ C \, =\,4\pi \, rh \, +\,20\pi \,r^2\]\)

We substitute the expression we found for \( h \) into this cost function:
\(\[ C \, =\,4\pi \,r \,\left(\,\frac{1000 \, -\,x\frac{4}{3}\pi \,r^3}{\pi \,r^2} \,\right) \, + \,20\pi \,r^2\]\)

Simplifying this expression gives us:
\(\[ C \, =\,x\frac{4000}{r} \, +\,\left(\,\frac{16}{3}\pi^2 \,r^3\,\right)\]\)

This cost function helps us find the optimal dimensions for the tank.

Derivatives and Critical Points

To find the radius that minimizes the cost, we need to find the critical points of the cost function by differentiating and solving for \(r\). The derivative of the cost function \( C \) with respect to \( r \) is:
\(\[ C' \, =\,-x\frac{4000}{r^2} \, +\,\left(\,\frac{16}{3}\,\right) \,3\pi^2 \,r^2\]\)

Simplify this to:
\(\[ C' \, =\,x-x\frac{4000}{r^2} \, +\,16\pi^2 \,r^2\]\)

To find the critical points, we set the derivative equal to zero and solve for \( r \):
\(\[ -\frac{4000}{r^2} \, +\,16\pi^2 \,r^2 \, =\,0\]\)

Solving for \( r \), we get:
\(\[ 16\pi^2 \,r^4 \, =\,4000\]\)

\(\[ r^4 \, =\,\frac{4000}{16\pi^2}\]\)

\(\[ r\, =\,\left(\frac{250}{\pi^2}\,\right)^{1/4} \, \approx\,3.63 \,\text{feet}\]\)

Next, we calculate \( h \) by substituting \( r \approx 3.63 \) feet back into the equation for \( h \):
\(\[ h \, =\,\frac{1000 \ \, -\,\frac{4}{3} \,\pi \, (3.63)^3}{\pi \, (3.63)^2} \, \approx\, 7.25 \text{ feet}\]\)

Finally, substitute these values into the cost function to determine the minimum cost:
\(\[ C \, =\,4\pi \, (3.63) \, (7.25) \, \ +\, 20\pi \, (3.63)^2\]\)
\(\[ \approx\,885.07 \, +\, 829.38 \, \approx\, 1714.45 \,dollars\]\)

The critical points involving derivatives ensure we've found the minimal cost for constructing the tank.