A can in the shape of a right circular cylinder is to be constructed to contain 1000 cm$^3$. The circular top and bottom of the can must have a radius of 0.25 cm more than the radius of the can so that the excess can be used to form a seal with the side. The sheet of material being formed into the side of the can must also be 0.25 cm longer than the circumference of the can so that a seal can be formed. Find, to within $10^{-4}$, the minimum amount of material needed to construct the can.

I'm at a loss as to what to do here. I'm not sure how to work this problem, as I can't find any discussion of optimization problems in my textbook. Can someone show me how to optimize this problem? The chapter it appears in is on zeros of polynomials and Muller's method.


Hint: First, think about what you are given and what variables express the problem. Let $r$ be the radius of the can, $h$ its height. You are given that the volume, $\pi r^2 h=1000$. The area of the material is then $2\pi (r+.25)^2 + (2 \pi r+0.25)h$. We are asked to choose either $r$ or $h$ to minimize the area. Take the volume equation and find $h=\frac {1000}{\pi r^2}$. Insert that into the area equation and you will get area as a function of $r$. Take the derivative and set to zero. The question indicates you will get something you are not expected to solve analytically. Instead, you should do a numeric solution and you are given the acceptable error.

  • 1
    $\begingroup$ Area of top and bottom? Additional 0.25 cm on circumference ? $\endgroup$ – DJohnM Mar 3 '13 at 6:37
  • $\begingroup$ Thank you! I knew the basic method for optimization (using the derivative), but I was looking for something completely different. I didn't think to try and solve using the derivative and then use another root-finding method. $\endgroup$ – SSumner Mar 3 '13 at 16:12
  • $\begingroup$ @user: missed that. Fixed. Thanks $\endgroup$ – Ross Millikan Mar 3 '13 at 17:08
  • $\begingroup$ $2\pi (r+.25)^2$ for the top and bottom? $\endgroup$ – DJohnM Mar 3 '13 at 18:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.