Math Problem Statement

What are the dimensions of an aluminum can that holds 320in^3 of juice and uses the least material?

Solution

To determine the dimensions of an aluminum can that holds 320 in³ of juice and uses the least material, we need to minimize the surface area of the can while ensuring that it holds the given volume.

Assumptions:

  • The can is cylindrical, which is a common shape for beverage cans.

  • The formula for the volume VV of a cylinder is: V=πr2hV = \pi r^2 h where rr is the radius of the base and hh is the height.

  • The formula for the surface area AA of a cylinder (including the top, bottom, and side) is: A=2πr2+2πrhA = 2\pi r^2 + 2\pi rh where the first term represents the areas of the top and bottom circles, and the second term is the lateral surface area.

Given:

  • Volume V=320in3V = 320 \, \text{in}^3

We need to minimize AA, the surface area, for the fixed volume. So, we have the following steps:

Step 1: Express height in terms of radius

Using the volume equation, solve for hh: 320=πr2h320 = \pi r^2 h h=320πr2h = \frac{320}{\pi r^2}

Step 2: Substitute hh into the surface area formula

Now, substitute h=320πr2h = \frac{320}{\pi r^2} into the surface area equation: A=2πr2+2πr(320πr2)A = 2\pi r^2 + 2\pi r \left( \frac{320}{\pi r^2} \right) Simplify the second term: A=2πr2+640rA = 2\pi r^2 + \frac{640}{r}

Step 3: Minimize the surface area

To minimize the surface area, take the derivative of AA with respect to rr and set it equal to 0: dAdr=4πr640r2\frac{dA}{dr} = 4\pi r - \frac{640}{r^2} Set this equal to 0: 4πr=640r24\pi r = \frac{640}{r^2} Multiply both sides by r2r^2: 4πr3=6404\pi r^3 = 640 Solve for rr: r3=6404πr^3 = \frac{640}{4\pi} r3=160πr^3 = \frac{160}{\pi} r=(160π)13r = \left( \frac{160}{\pi} \right)^{\frac{1}{3}} Using a calculator to find the value: r4.35inr \approx 4.35 \, \text{in}

Step 4: Calculate the height

Now that we have the radius rr, substitute it back into the equation for height: h=320πr2h = \frac{320}{\pi r^2} h=320π(4.35)2h = \frac{320}{\pi (4.35)^2} h5.37inh \approx 5.37 \, \text{in}

Final Dimensions:

  • Radius r4.35inr \approx 4.35 \, \text{in}
  • Height h5.37inh \approx 5.37 \, \text{in}

Conclusion:

The optimal dimensions for the aluminum can that holds 320 in³ of juice and uses the least material are approximately:

  • Radius: 4.35 inches
  • Height: 5.37 inches

Would you like more details or have any questions about this solution?

Related Questions:

  1. What is the process for deriving the surface area formula for a cylinder?
  2. How do you find the volume of different 3D shapes like spheres or cones?
  3. Why is minimizing surface area important in packaging design?
  4. How does the material used in packaging affect the total cost of manufacturing?
  5. What is the relationship between volume and surface area in optimization problems?

Tip:

When solving optimization problems, always begin by expressing the variables you're trying to optimize in terms of each other using the given constraints. This simplifies the problem and leads to a solution by calculus!

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Math Problem Analysis

Mathematical Concepts

Optimization
Calculus
Geometry
Volume of a cylinder
Surface area of a cylinder

Formulas

Volume of a cylinder: V = πr²h
Surface area of a cylinder: A = 2πr² + 2πrh

Theorems

Optimization using calculus
Minimization by taking derivatives

Suitable Grade Level

Grades 11-12