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.
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.
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
To solve this problem, let's break it down step by step:
1) Express the cost of the side and the two bases in terms of r, h, and k:
The area of the side of the cylinder is given by the formula A_side = 2πrh. Since the cost of the side material is k cents per cm square, the cost of the side is given by C_side = k * A_side = k * 2πrh.
The area of each base of the cylinder is given by the formula A_base = πr^2. Since the cost of the base material is twice that of the side, the cost of each base is given by C_base = 2k * A_base = 2k * πr^2.
2) Express the total cost of manufacturing this can in terms of r only:
The total cost is the sum of the cost of the side and the cost of the two bases:
C_total = C_side + 2 * C_base
= k * 2πrh + 2 * (2k * πr^2)
= 2πkrh + 4kπr^2
3) Find r that minimizes the cost:
To minimize the cost, we need to find the derivative of the cost function with respect to r, and set it equal to zero.
dC_total/dr = 4πkrh + 8kπr = 0
Simplifying the equation:
4πkrh + 8kπr = 0
r(4πkh + 8πr) = 0
Since r cannot be zero, we can divide both sides by r:
4πkh + 8πr = 0
4kh + 8r = 0
8r = -4kh
r = -0.5kh
Since we are looking for a positive radius, we ignore the negative sign and obtain the optimal value for r:
r = 0.5kh
4) Deduce the corresponding h:
To find the corresponding h, we substitute the value of r obtained in step 3 back into the equation for the volume of a cylinder:
V = πr^2h
Substituting r = 0.5kh:
100 = π(0.5kh)^2h
100 = π(0.25k^2h^3)
h^3 = (100 / (0.25πk^2))
Taking the cube root of both sides:
h = (100 / (0.25πk^2))^(1/3)
Thus, the corresponding h is given by h = (100 / (0.25πk^2))^(1/3).