Help Solve Calculus Project - Economical Can Dimensions

[hankjr]

I was assined a pop can project in calculus class that contained three parts. I did the first 2 parts with little to no trouble but I really need help on the third because I don't even know where to start. Plz help me as soon as possible!

C.) When a 12 oz can is to be manufactured there is no waste involved in cutting the metal that makes the vertical sides of the can ( why?), but the top and bottom are made from square pieces of metal and the corners of the square are wasted. Find the dimensions of the most economical can ( remember that the square corner pieces are not used but would be part of the materials purchased to make the can.

PLEASE PLEASE HELP ME , I NEED TO FINISH SOON !

 

[OlderDan]

I had some trouble interpreting this, but I think all they are looking for is the minimum area of metal needed to make a 12 oz cylindrical can, subject to the fact that the circular ends of diameter d and raduis r require excess material. The total area of metal would be A = pi*d*h + 2d², with the volume being V = h*pi*r²

 

[HallsofIvy]

I was assined a pop can project in calculus class that contained three parts. I did the first 2 parts with little to no trouble but I really need help on the third because I don't even know where to start. Plz help me as soon as possible!

C.) When a 12 oz can is to be manufactured there is no waste involved in cutting the metal that makes the vertical sides of the can ( why?), but the top and bottom are made from square pieces of metal and the corners of the square are wasted. Find the dimensions of the most economical can ( remember that the square corner pieces are not used but would be part of the materials purchased to make the can.

PLEASE PLEASE HELP ME , I NEED TO FINISH SOON !

There is no waste from the vertical sides because that is a rectangle shaped into a cylinder. Let h be the height of the cylinder, x be the length of the sides of the square from which the top and bottom are cut. I hope it is obvious that x should be the diameter of the the top and bottom (otherwise we could first cut a square to that size and then cut the circle). The radius of the top and bottom is x/2 so the volume of the cylinder is $\pi (x/2)^2 h= \frac{\pi}{8}x^2 h= 12$ (12 oz is a volume.) The waste material will be the area of the square minus the area of the circle: $x^2- \pi/4 x^2= (1- \pi/4)x^2$. You want to minimize that subject to the condition that $\pi (x/2)^2 h= \frac{\pi}{8}x^2 h= 12$.

 

[OlderDan]

There is no waste from the vertical sides because that is a rectangle shaped into a cylinder. Let h be the height of the cylinder, x be the length of the sides of the square from which the top and bottom are cut. I hope it is obvious that x should be the diameter of the the top and bottom (otherwise we could first cut a square to that size and then cut the circle). The radius of the top and bottom is x/2 so the volume of the cylinder is $\pi (x/2)^2 h= \frac{\pi}{8}x^2 h= 12$ (12 oz is a volume.) The waste material will be the area of the square minus the area of the circle: $x^2- \pi/4 x^2= (1- \pi/4)x^2$. You want to minimize that subject to the condition that $\pi (x/2)^2 h= \frac{\pi}{8}x^2 h= 12$.

I think minimizing the waste is not sufficient. x can be made arbitrarily small by making h as big as it needs to be to form a 12 oz can with tiny circles for ends. The total area of metal used to make the can, including the waste, must be minimized.

 

[HallsofIvy]

Ah, you are right. I misread the problem. The problem is to minimize $\pi x h+ x^2$ subject to the condition $\frac{\pi}{8}x^2h= 12$.