Math Problem Statement

An open-top, hollow plastic cylinder has a volume of 162 cm^3. Find the minimum surface area of the cylinder.

Solution

We are given an open-top hollow cylinder, with a volume of 162cm3162 \, \text{cm}^3, and we need to find the minimum surface area.

Formulae involved:

  1. Volume of a cylinder: V=πr2hV = \pi r^2 h where:

    • VV is the volume,
    • rr is the radius,
    • hh is the height.
  2. Surface area of an open-top cylinder: A=πr2+2πrhA = \pi r^2 + 2\pi r h where:

    • The first term πr2\pi r^2 is the area of the base,
    • The second term 2πrh2\pi r h is the lateral surface area.

Step 1: Express height hh in terms of rr and VV

From the volume formula: V=πr2h    h=Vπr2V = \pi r^2 h \implies h = \frac{V}{\pi r^2} Substitute V=162cm3V = 162 \, \text{cm}^3: h=162πr2h = \frac{162}{\pi r^2}

Step 2: Substitute hh into the surface area formula

A=πr2+2πr(162πr2)A = \pi r^2 + 2\pi r \left( \frac{162}{\pi r^2} \right) Simplify: A=πr2+324rA = \pi r^2 + \frac{324}{r}

Step 3: Minimize the surface area

To minimize AA, we take the derivative of AA with respect to rr, set it to zero, and solve for rr.

dAdr=2πr324r2\frac{dA}{dr} = 2\pi r - \frac{324}{r^2} Set dAdr=0\frac{dA}{dr} = 0: 2πr=324r22\pi r = \frac{324}{r^2} Multiply both sides by r2r^2: 2πr3=3242\pi r^3 = 324 Solve for rr: r3=3242π=162πr^3 = \frac{324}{2\pi} = \frac{162}{\pi} r=(162π)13r = \left( \frac{162}{\pi} \right)^{\frac{1}{3}}

Let me calculate the value of rr.The radius rr is approximately 3.72cm3.72 \, \text{cm}.

Step 4: Find the corresponding height hh

Using the formula h=162πr2h = \frac{162}{\pi r^2}, we can calculate the height. Let me compute that next.The height hh is also approximately 3.72cm3.72 \, \text{cm}.

Step 5: Calculate the minimum surface area

Now, let's calculate the surface area using the formula: A=πr2+324rA = \pi r^2 + \frac{324}{r} Let me compute the minimum surface area.The minimum surface area of the open-top hollow cylinder is approximately 131cm2131 \, \text{cm}^2.

Would you like further details or have any questions?

Here are 5 related questions you might consider:

  1. How does the surface area change with different cylinder heights?
  2. What would the surface area be if the cylinder had a closed top?
  3. How do you derive the optimal dimensions for a different volume?
  4. Can this method be applied to minimize the surface area of other shapes?
  5. How does the ratio between height and radius affect the efficiency of material usage?

Tip: When solving optimization problems involving geometry, always check if taking derivatives helps find minimum or maximum values efficiently.

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Math Problem Analysis

Mathematical Concepts

Optimization
Surface Area
Derivatives
Geometry

Formulas

Volume of cylinder: V = πr^2h
Surface area of open-top cylinder: A = πr^2 + 2πrh
Derivative to minimize surface area: dA/dr = 2πr - 324/r^2

Theorems

Optimization using derivatives
Surface area and volume relationships

Suitable Grade Level

Grades 10-12