Asked by zeina
A cylindrical can is to be made to hold 100cm cubic.The material for its top and bottom cost twice as much per cm square as that for its side.
The objective of this problem is to find the radius and the height of this cylinder such that its cost is minimized.Denote the radius of this can by r and its height by h. Let k be the cost,expressed in cents,of 1 cm square of the side of this can.
1)Express the cost of the side and that two bases in terms of r, h and k.
2)Express the total cost of manufacturing this can in terms of r only.
3)Find r that minimizes the cost.
4)Deduce the corresponding h.
The objective of this problem is to find the radius and the height of this cylinder such that its cost is minimized.Denote the radius of this can by r and its height by h. Let k be the cost,expressed in cents,of 1 cm square of the side of this can.
1)Express the cost of the side and that two bases in terms of r, h and k.
2)Express the total cost of manufacturing this can in terms of r only.
3)Find r that minimizes the cost.
4)Deduce the corresponding h.
Answers
Answered by
bobpursley
Let r be the radius, h the height.
Cost=2K(2PIr^2)+ k*h*PI*(2r)
I will be happy to critique your work on this. A most excellent problem.
Cost=2K(2PIr^2)+ k*h*PI*(2r)
I will be happy to critique your work on this. A most excellent problem.
Answered by
Reiny
we know volume = πr^2h = 100
h = 100/(πr^2)
k = 2(2πr^2) + 1(2πrh)
= 4πr^2 = 2πr(100/(πr^2))
= = 4πr^2 + 200/r
dk/dr = 8πr - 200/r^2 = 0 for a max/min of k
8πr = 200/r^2
r^3 = 200/(8π)
take it from here
h = 100/(πr^2)
k = 2(2πr^2) + 1(2πrh)
= 4πr^2 = 2πr(100/(πr^2))
= = 4πr^2 + 200/r
dk/dr = 8πr - 200/r^2 = 0 for a max/min of k
8πr = 200/r^2
r^3 = 200/(8π)
take it from here
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.