Optimizing Metal Cost for Cylindrical Can Construction

[courtrigrad]

Hello all

For this problem:

3. A manufacturer wishes to construct a cylindrical can to hold 100 pi in^3. The top and bottom of the can are to be stronger than the sides. The tin used in making the top and bottom will cost 2.5 cents per square inch while the metal used in making the sides will cost 1.35 cents per square inch. What dimensions should be used to minimize the cost of the metal?

pi*r^2 = 100pi
Area of sides = 2pi*r*h
Area of top and bottom: 2pi*r^2

h = 100/ r^2

2pi*r(100/r^2) + 2*pi*r^2

Now do I multiply the cost of the sides by the coefficients?

Any help is appreciated

Thanks!
courtrigrad is online now Edit/Delete Message

 

[HallsofIvy]

You have "pi r^2= 100 pi" when, of course, it should be pi h r^2= 100 pi. (You knew that since you then have h= 100/r^2).

Yes, the 'cost' is the area of each part times the cost of each part: The cost of the side will be 1.35(2pi r)(100/r^2) (which is 270pi/r) and the cost of the top and bottom will be (2.5)(2 pi r^2)= 5pi r^2. The cost of the entire can is the sum of those:
270pi/r+ 5pi r^2. Set the derivative of that equal to 0 to find the value of r that makes that a minimum. Don't forget to find h too!

 

[courtrigrad]

thanks a lot