Math Problem Statement

For its beef stew, Betty Moore Company uses aluminum containers that have the form of right circular cylinders. Find the radius and height of a container if it has a capacity of 32 in³ and is constructed using the least amount of metal. (Round your answers to two decimal places.)

Solution

The problem asks us to find the radius and height of a right circular cylinder that has a capacity (volume) of 32 in³ and is constructed using the least amount of metal, meaning it aims to minimize the surface area.

Key formulas involved:

  1. Volume of a cylinder:
    V=πr2hV = \pi r^2 h where rr is the radius, hh is the height, and V=32in3V = 32 \, \text{in}^3.

  2. Surface area of a cylinder (the amount of metal used):
    A=2πr2+2πrhA = 2\pi r^2 + 2\pi r h where the first term represents the area of the two circular ends, and the second term is the area of the curved side.

Steps to solve:

  1. Use the volume equation to express height hh in terms of radius rr:
    From the volume equation: h=Vπr2=32πr2h = \frac{V}{\pi r^2} = \frac{32}{\pi r^2}

  2. Substitute this into the surface area formula:
    A=2πr2+2πr(32πr2)=2πr2+64rA = 2\pi r^2 + 2\pi r \left( \frac{32}{\pi r^2} \right) = 2\pi r^2 + \frac{64}{r}

  3. Minimize the surface area by taking the derivative of AA with respect to rr and setting it equal to 0: dAdr=4πr64r2\frac{dA}{dr} = 4\pi r - \frac{64}{r^2} Set this equal to zero to find the critical points: 4πr=64r24\pi r = \frac{64}{r^2} Multiply both sides by r2r^2: 4πr3=644\pi r^3 = 64 Solve for rr: r3=644π=16πr^3 = \frac{64}{4\pi} = \frac{16}{\pi} r=16π3r = \sqrt[3]{\frac{16}{\pi}}

  4. Find the corresponding height hh: Once we have rr, substitute it back into the equation for height: h=32πr2h = \frac{32}{\pi r^2}

Let me compute the exact values for you.The radius and height of the container that minimizes the amount of metal are approximately:

  • Radius: 1.72in1.72 \, \text{in}
  • Height: 3.44in3.44 \, \text{in}

Would you like further details or have any questions?

Here are some related questions to explore:

  1. What is the total surface area of the cylinder with these dimensions?
  2. How does changing the volume affect the optimal dimensions of the cylinder?
  3. Could the container's shape become more cost-effective with a different shape than a cylinder?
  4. How is the optimization process affected if we aim to minimize volume instead of surface area?
  5. How would the dimensions change if the container is made for a different material with specific constraints?

Tip: When optimizing for cost, always consider how changing one dimension affects the others through constraints like volume.

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Math Problem Analysis

Mathematical Concepts

Optimization
Calculus
Geometry
Volume and Surface Area

Formulas

Volume of a cylinder: V = πr²h
Surface area of a cylinder: A = 2πr² + 2πrh
Optimization through derivatives: dA/dr = 4πr - 64/r²

Theorems

Optimization using calculus

Suitable Grade Level

Grades 11-12