Math Problem Statement

A cylindrical can, open at the top, is to hold 650 cm³ of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can. Enter your answer with rational exponents, and use pi to represent π.

Solution

To solve this problem, we need to determine the radius rr and height hh of a cylindrical can with a volume of 650 cm³, that minimizes the surface area of the can (since this will minimize the material needed). The can is open at the top, meaning the surface area consists only of the bottom circle and the cylindrical side.

Step 1: Define the Volume of the Cylinder

The volume VV of a cylinder is given by: V=πr2hV = \pi r^2 h We're told the volume is 650 cm³, so: πr2h=650\pi r^2 h = 650 This can be rearranged to express hh in terms of rr: h=650πr2h = \frac{650}{\pi r^2}

Step 2: Define the Surface Area to be Minimized

The surface area AA of an open-top cylinder is the sum of the area of the bottom and the lateral surface area: A=πr2+2πrhA = \pi r^2 + 2 \pi r h Substitute hh from the volume equation: A=πr2+2πr(650πr2)A = \pi r^2 + 2 \pi r \left( \frac{650}{\pi r^2} \right) Simplify the equation: A=πr2+1300rA = \pi r^2 + \frac{1300}{r}

Step 3: Find the Minimum Surface Area

To find the value of rr that minimizes AA, take the derivative of AA with respect to rr and set it to zero: dAdr=2πr1300r2\frac{dA}{dr} = 2\pi r - \frac{1300}{r^2} Set dAdr=0\frac{dA}{dr} = 0: 2πr=1300r22\pi r = \frac{1300}{r^2} Multiply both sides by r2r^2: 2πr3=13002\pi r^3 = 1300 Solve for rr: r3=13002π=650πr^3 = \frac{1300}{2\pi} = \frac{650}{\pi} r=(650π)13r = \left( \frac{650}{\pi} \right)^{\frac{1}{3}}

Step 4: Calculate hh

Substitute r=(650π)13r = \left( \frac{650}{\pi} \right)^{\frac{1}{3}} back into the equation for hh: h=650πr2=650π(650π)23h = \frac{650}{\pi r^2} = \frac{650}{\pi \left( \frac{650}{\pi} \right)^{\frac{2}{3}}} Simplify: h=(650π)13h = \left( \frac{650}{\pi} \right)^{\frac{1}{3}}

Final Answer

So the values for rr and hh that minimize the surface area are: Radius r=(650π)13\text{Radius } r = \left( \frac{650}{\pi} \right)^{\frac{1}{3}} Height h=(650π)13\text{Height } h = \left( \frac{650}{\pi} \right)^{\frac{1}{3}}

Would you like further explanation on any part of this solution?

Follow-up Questions

  1. How would the formula change if the can were closed on top?
  2. What would the effect be on rr and hh if the volume were doubled?
  3. How can we verify if this solution gives a minimum or maximum surface area?
  4. What is the approximate decimal value of (650π)13\left( \frac{650}{\pi} \right)^{\frac{1}{3}}?
  5. How does this approach apply to other shapes with fixed volumes?

Tip

When working with optimization problems, always define your constraints first and express one variable in terms of the others to reduce complexity.

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Math Problem Analysis

Mathematical Concepts

Optimization
Calculus
Geometry of Solids
Cylinders

Formulas

Volume of a cylinder: V = πr²h
Surface area of an open-top cylinder: A = πr² + 2πrh
Derivative for optimization: dA/dr = 0

Theorems

Optimization by differentiation

Suitable Grade Level

Grades 11-12