Math Problem Statement

please solve

A cylindrical beverage can is to be produced in such a way that as little sheet metal as possible is used. What is the height and diameter of an optimal beverage can that has a capacity of 0.5 litres?

Solution

To solve the optimization problem for the cylindrical beverage can:

Problem Breakdown:

  • Objective: Minimize the surface area of the cylinder (to minimize sheet metal used).
  • Constraint: The cylinder must have a volume of 0.5 liters, i.e., V=0.5L=500cm3V = 0.5 \, \text{L} = 500 \, \text{cm}^3 (converting to cm³).
  • Unknowns: Height hh and radius rr of the cylinder.

Step 1: Expressions for Volume and Surface Area

  1. Volume: V=πr2h=500V = \pi r^2 h = 500 Solve for hh: h=500πr2h = \frac{500}{\pi r^2}

  2. Surface Area: The surface area includes the side and two circular ends: S=2πrh+2πr2S = 2\pi r h + 2\pi r^2 Substitute h=500πr2h = \frac{500}{\pi r^2} into SS: S=2πr(500πr2)+2πr2S = 2\pi r \left(\frac{500}{\pi r^2}\right) + 2\pi r^2 Simplify: S=1000r+2πr2S = \frac{1000}{r} + 2\pi r^2


Step 2: Minimize Surface Area

To minimize SS, take the derivative dSdr\frac{dS}{dr} and set it to zero.

  1. Differentiate: dSdr=1000r2+4πr\frac{dS}{dr} = -\frac{1000}{r^2} + 4\pi r

  2. Set dSdr=0\frac{dS}{dr} = 0: 1000r2+4πr=0-\frac{1000}{r^2} + 4\pi r = 0 Rearrange: 1000r2=4πr\frac{1000}{r^2} = 4\pi r Multiply through by r2r^2: 1000=4πr31000 = 4\pi r^3 Solve for r3r^3: r3=10004π=250πr^3 = \frac{1000}{4\pi} = \frac{250}{\pi} Take the cube root: r=250π3r = \sqrt[3]{\frac{250}{\pi}}

  3. Approximate rr (using π3.1416\pi \approx 3.1416): r79.577534.31cmr \approx \sqrt[3]{79.5775} \approx 4.31 \, \text{cm}


Step 3: Calculate hh

Using h=500πr2h = \frac{500}{\pi r^2}: h=500π(4.31)2h = \frac{500}{\pi (4.31)^2} h5003.141618.5750058.358.57cmh \approx \frac{500}{3.1416 \cdot 18.57} \approx \frac{500}{58.35} \approx 8.57 \, \text{cm}


Final Answer:

  • Radius: r4.31cmr \approx 4.31 \, \text{cm}
  • Height: h8.57cmh \approx 8.57 \, \text{cm}

Would you like detailed steps for approximations or further clarifications?

Related Questions:

  1. What happens to the dimensions if the volume is increased to 1 liter?
  2. How does the shape of the can affect material costs in production?
  3. Can this optimization process be extended to include thermal insulation requirements?
  4. How does using a different material for the can influence the optimization constraints?
  5. What are the trade-offs between minimizing surface area and maximizing stackability?

Tip:

Always convert all measurements to the same unit (e.g., cm³) before solving optimization problems.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Optimization
Geometry
Differentiation

Formulas

Volume of a cylinder: V = πr^2h
Surface area of a cylinder: S = 2πrh + 2πr^2
Derivative of the surface area with respect to radius: dS/dr = -1000/r^2 + 4πr

Theorems

Critical point analysis to minimize a function using derivatives

Suitable Grade Level

Grades 11-12