A cylindrical container is to hold 20π cm^3. The top and bottom are made of a material that costs $0.40 per cm^2. The material for the curved side costs $0.32 per cm^2.

Find the height in centimeters of the most economical container.

let the radius be r

let the height be h

V = πr^2 h
πr^2 h = 20π
h = 20/r^2

Cost = .40(2πr^2) + .32(2πrh)
C = .8πr^2 + .64πr(20/r^2)
= .8πr^2 + 12.8π/r
dC/dr = 1.6πr - 12.8π/r^2
= 0 for a min of C

1.6πr = 12.8π/r^2
1.6r^3 = 12.8
r^3 = 8
r = 2
then h = 20/4 = 5

the cylinder should be 5 cm high and have a radius of 2 cm.

To find the height of the most economical container, we first need to determine the total cost associated with the different components of the container.

Let's denote the radius of the cylindrical container as 'r' and the height as 'h'. Given that the volume of the container is 20π cm^3, we can express it as follows:

Volume of a cylinder = πr^2h = 20π

Now, we need to minimize the cost, which consists of the cost of the top and bottom covers as well as the cost of the curved side. The cost of the flat top and bottom covers is calculated by multiplying the area by the cost per cm^2, while the cost of the curved side is calculated by multiplying the curved surface area by the cost per cm^2.

The area of a circle is given by A = πr^2, and the curved surface area of a cylinder is given by CSA = 2πrh.

Let's compute the total cost:

Cost = Cost of top and bottom covers + Cost of curved side
= 2(πr^2)(0.40) + (2πrh)(0.32)
= 0.80πr^2 + 0.64πrh

Now, we need to find the height that minimizes this cost. We can achieve this by finding the critical points of the cost function with respect to 'h'.

First, let's differentiate the cost function with respect to 'h' while treating 'r' as a constant:

dCost/dh = 0.64πr

To find the critical points, set this derivative equal to zero:

0.64πr = 0

Since πr ≠ 0 (the radius cannot be zero), there are no critical points with respect to 'h'.

Next, we know that the height of the container should be positive, so we take the limit of the cost function as 'h' approaches positive infinity:

lim(h→∞) Cost = ∞

The cost increases without bound as the height increases. Hence, to minimize the cost, the height of the container should be infinite, which is not practical.

Since there are no critical points with respect to 'h' and taking the limit as 'h' approaches infinity yields an impractical result, we conclude that there is no minimum cost for a cylindrical container with a fixed volume. In other words, the cost depends on the dimensions you choose.

Therefore, we cannot find the height of the most economical container without additional information or constraints.