Math Problem Statement

A cylindrical can has a volume of 686 pi cm cubed. What dimensions yield the minimum surface​ area? Question content area bottom Part 1 The radius of the can with the minimum surface area is    enter your response here cm. ​(Simplify your​ answer.) Part 2 The height of the can with the minimum surface area is    enter your response here cm. ​(Simplify your​ answer.)

Solution

To solve for the dimensions that yield the minimum surface area of the cylindrical can, we use the following steps:

Part 1: The Radius of the Can with Minimum Surface Area

Given the volume V=686πcm3V = 686 \pi \, \text{cm}^3, the volume of a cylinder is calculated by: V=πr2hV = \pi r^2 h where rr is the radius and hh is the height.

From this equation, we can solve for hh in terms of rr: h=Vπr2=686ππr2=686r2h = \frac{V}{\pi r^2} = \frac{686 \pi}{\pi r^2} = \frac{686}{r^2}

Next, the surface area SS of a cylinder (including the top and bottom) is given by: S=2πr2+2πrhS = 2\pi r^2 + 2\pi r h Substitute h=686r2h = \frac{686}{r^2} into the surface area formula: S=2πr2+2πr686r2S = 2\pi r^2 + 2\pi r \cdot \frac{686}{r^2} Simplify the expression: S=2πr2+1372πrS = 2\pi r^2 + \frac{1372\pi}{r}

To minimize the surface area, we need to take the derivative of SS with respect to rr and set it equal to zero: dSdr=4πr1372πr2\frac{dS}{dr} = 4\pi r - \frac{1372\pi}{r^2} Set this equal to zero: 4πr1372πr2=04\pi r - \frac{1372\pi}{r^2} = 0 Divide both sides by π\pi: 4r1372r2=04r - \frac{1372}{r^2} = 0 Multiply both sides by r2r^2 to eliminate the fraction: 4r3=13724r^3 = 1372 Solve for rr: r3=13724=343r^3 = \frac{1372}{4} = 343 r=3433=7cmr = \sqrt[3]{343} = 7 \, \text{cm}

Part 2: The Height of the Can with Minimum Surface Area

Now that we have the radius, r=7cmr = 7 \, \text{cm}, substitute it into the equation for height: h=686r2=68672=68649=14cmh = \frac{686}{r^2} = \frac{686}{7^2} = \frac{686}{49} = 14 \, \text{cm}

Final Answer:

  • Radius: 7cm7 \, \text{cm}
  • Height: 14cm14 \, \text{cm}

Would you like more details on any part of this process?

Here are 5 questions to expand on this:

  1. What would happen to the surface area if the volume were doubled?
  2. Can you derive the formula for the surface area of a cylinder from basic principles?
  3. What is the significance of the derivative in finding the minimum surface area?
  4. How does the ratio of height to radius affect the surface area?
  5. Can this optimization technique be applied to other shapes?

Tip: Optimization problems often involve taking derivatives to find minimum or maximum values by identifying critical points.

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Math Problem Analysis

Mathematical Concepts

Geometry
Optimization
Derivatives
Cylindrical Volume and Surface Area

Formulas

Volume of a cylinder: V = πr²h
Surface area of a cylinder: S = 2πr² + 2πrh
Derivative of surface area with respect to radius: dS/dr = 4πr - 1372π/r²

Theorems

Critical points found using derivatives
Optimization of functions using calculus

Suitable Grade Level

Grades 10-12