# Thread: applied max and min word problem

1. ## applied max and min word problem

I am not getting this type of problem.
The question is: A soup can in the shape of a right circular cylinder of radius r and height h is to have a prescribed volume V. The top and bottom are cut from squares as shown in the figure (the figure is a circle inscribed in a square). If the shaded corners (the corners of the square outside of the circle) are wasted, but there is no other waste, find the ratio r/h for the can requiring the least material, including waste.

In my book, it has steps for solving these problems. It says to draw a picture, then find a formula for the quantity to be maximized or minimized. Then using the conditions stated in the problem to eliminate variables, express the quantity as a function of one variable. Then find the interval of possible values from physical restrictions and solve.

I know the formulas for volume and surface area of a cylinder: V=$\displaystyle \pi r^2$h and S=2$\displaystyle \pi$rh (that is the equation for surface area, right?). But I'm not sure exactly how to use them here.
I know you need to get an equation with one variable in it, either r or h, and then take the derivative, but I don't know how r and h relate to each other in this problem.

2. Hi, and welcome to the forum.

I don't know how r and h relate to each other in this problem.
You are told that the can has to have a prescribed volume V. Using this fact, you can express, say, h as a function of an unknown r and a known constant V.

After that, find the total area of metal that is needed to make a can of radius r and height h (note that the top and bottom are squares, not circles). Substitute h from the previous paragraph to get a function of r only. Using differentiation, find r in terms of V that gives the minimum of the function.

Now you should have an equality on r and V. Substituting $\displaystyle V=\pi r^2h$, you get an equation on r and h, from where you can get the required ratio r/h.