SOLUTION: 2. A cylindrical can is to hold 16πcm3 of mango juice. The cost per square meter of constructing the top and bottom is twice the cost per square meter of constructing the card

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Question 975337: 2. A cylindrical can is to hold 16πcm3 of mango juice. The cost per square meter of constructing the top and bottom is twice the cost per square meter of constructing the cardboard side. What dimensions minimize the total cost of construction?
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A cylindrical can is to hold 16πcm3 of mango juice. The cost per square meter of constructing the top and bottom is twice the cost per square meter of constructing the cardboard side. What dimensions minimize the total cost of construction?
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Area of top + bottom = 2pi%2Ar%5E2
Lateral area = 2pi%2Ar%2Ah
Vol+=+pi%2Ar%5E2h+=+16pi
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Cost+=+k%2A%282pi%2Ar%2Ah+%2B+4pir%5E2%29 where k is some constant
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Vol+=+pi%2Ar%5E2h+=+16pi
r%5E2h+=+16
h+=+16%2Fr%5E2
Sub for h in the cost eqn:
C+=+k%2A%2832pi%2Fr+%2B+4pir%5E2%29
Find the 1st derivative of Cost wrt r, set = 0:
dC/dr = k%2A%28-32pi%2Fr%5E2+%2B+8pi%2Ar%29+=+0
-32%2Fr%5E2+%2B+8r+=+0
r+-+4%2Fr%5E2+=+0
r%5E3+-+4+=+0
r+=+root%283%2C4%29


Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The volume of a cylinder is given by , so




hence



The area of either the top or bottom of the can is given by

,

but there are two ends, so the area of the ends is



The area of the side of the can is the circumference of the end times the height of the can, so:



The total surface area of the can is then:



But if we substitute we can create a single variable function for the total surface area:




Using the fact that the ends cost twice as much per unit area than the side, we can now write a function for the cost in terms of the height:



Then take the first derivative:



Set the first derivative equal to zero and solve for the value of that yields a local extrema for the function:





Now take the 2nd derivative:



Which is positive for , hence

Now that we have a height value that minimizes the cost function, we can calculate the resulting radius value using the volume relationship.







John

My calculator said it, I believe it, that settles it