you are designing a 1000-cm^3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up by the can will therefore be: A=8r^2 +2(pi)(r)(h)
What is the ratio of h to r for the most economical can?
Never mind... I got the answer!
To find the ratio of h to r for the most economical can, we need to minimize the amount of aluminum used in the can. The equation given, A = 8r^2 + 2πrh, represents the total amount of aluminum used.
First, let's simplify the equation by factoring out common terms:
A = 2r(4r + πh)
To find the minimum amount of aluminum used, we can find the critical points of A with respect to h:
1. Differentiate A with respect to h:
dA/dh = 2rπ
2. Set dA/dh equal to zero to find the critical point:
2rπ = 0
Since π (pi) is a non-zero constant, we can conclude that there are no critical points for A with respect to h. Therefore, there is no minimum or maximum value for A with respect to h.
This means that there is no specific ratio of h to r that will result in the most economical can. Instead, the ratio can be any value as long as it satisfies the given constraints.
If you have any further questions, feel free to ask!