This is the question: If the cost of material to make the can in Example 3 is 5 cents per square inch for the top and bottom and 3 cents per square inch for the sides, what dimensions shoud be used to minimize the cost of making the can? [The answer is not the same as in Example 3.]
Example 3 is a sheet of metal (ch) rolled into a can with a top and bottom. The can's volume is 58 cubic inches. So the surface area equation is "c" or the circumfrence (2pi r) times height or "h." "H" is 58/pi r square. Then, I added a top and bottom, which is 2 pi r square. Added together and simplified, y= 116/x + 2 pi r squared. (This is also in the book, so I know it is correct.)
Per the books answers, the coordinates are (2.09773, 82.946845). These coordinates are found using the graphical minimum finder. These are the answers they say are NOT correct for this question.
But, since the material for the top is more expensive that that of the sides, wouldn't we want the top and bottom to have as small a radius as possible? I do not understand how to answer the question.
Any help is appreciated.
Mdd
Example 3 is a sheet of metal (ch) rolled into a can with a top and bottom. The can's volume is 58 cubic inches. So the surface area equation is "c" or the circumfrence (2pi r) times height or "h." "H" is 58/pi r square. Then, I added a top and bottom, which is 2 pi r square. Added together and simplified, y= 116/x + 2 pi r squared. (This is also in the book, so I know it is correct.)
Per the books answers, the coordinates are (2.09773, 82.946845). These coordinates are found using the graphical minimum finder. These are the answers they say are NOT correct for this question.
But, since the material for the top is more expensive that that of the sides, wouldn't we want the top and bottom to have as small a radius as possible? I do not understand how to answer the question.
Any help is appreciated.
Mdd
