Optimization problem
A cylindrical tin can is to be manufactured so that it will hold a specific volume V. If the materials for the ends of the can are twice as expensive as materials for the sides, what shape of can is most economical to manufacture? Find h in terms of r. Let the cost=$c per cm2, h is the height of the cylinder and r is the radius of the cylinder. You can assume that the cost is proportional to the surface area since the materials for a tin can have uniform thickness. Ignore the costs of forming the can, which are about the same for cans of any size
answer is h=4r
Answer
- The questioner was satisfied and accepted the answer, or
- The answer was disputed, but the judge evaluated it as 100% correct.

-
Leave a comment if you need any clarifications.
-
Sorry, there was a small mistake. It is fixed now. Please double check and let me know if you have any questions.
-
Since the question is tagged as a calculus question, I used calculus techniques to solve it. This problem could also be solved using Lagrange Multipliers in multivariable calculus.
-
@James Oxenham: Please review the revised solution and let us know if you find it satisfactory.
- answered
- 411 views
- $2.00
Related Questions
- Explain partial derivatives v3
- Applied calc question 2 and 3
- Use the equation to show the maximum, minimum, and minimum in the future.
- Proving f is continuous
- Explain why does gradient vector points in the direction of the steepest increase?
- Find the limit as x --> +inf
- Calc limit problem
- Find $\lim\limits _{n\rightarrow \infty} n^2 \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}$
The bounty is a bit low.