I'm a little stuck on this:
Suppose you are a food can manufacturer and have to supply circular cylindrical, sheet metal cans
of volume V = 256π cm3. The cost of the circular ends (top and bottom) per unit area is twice the
cost of the side per unit area. What are dimensions of the most economical can in terms of the
materials (ignoring the cost of attaching the pieces).
Okay so I know one of the sides is a rectangle of xh, and the two circles on top are 2pi*r^2
so you have 2pi*r^2 + xh, rearrange to h=(-2pi*r^2)/x, then plug it back in..? Am I doing this right, what is your final answer?
Thank you
Suppose you are a food can manufacturer and have to supply circular cylindrical, sheet metal cans
of volume V = 256π cm3. The cost of the circular ends (top and bottom) per unit area is twice the
cost of the side per unit area. What are dimensions of the most economical can in terms of the
materials (ignoring the cost of attaching the pieces).
Okay so I know one of the sides is a rectangle of xh, and the two circles on top are 2pi*r^2
so you have 2pi*r^2 + xh, rearrange to h=(-2pi*r^2)/x, then plug it back in..? Am I doing this right, what is your final answer?
Thank you
-
Volume = r^2 * pi * h = 256
h = 256/(pi * r^2)
The combined area of the circles is 2 * pi * r^2
The area of the side is the circumference of the circle times the height, or 2r * pi * h = 2r * pi * 256 / (pi * r^2) = 512/r
The cost per unit area of the side is some constant C; the cost per unit area of the circles is 2C.
So the total cost is 2C(2 pi r^2) + C (512/r) = 4C pi r^2 + 512C/r
The derivative of the cost is 8C pi r + -512C / r^2; set equal to zero and solve for r.
8C pi r + -512C / r^2 = 0
8C pi r = 512C / r^2
8C pi r^3 = 512C
r^3 = 64 / pi
r = 4 / pi^(1/3)
h = 256/(pi * r^2) = 256 / (pi * (4 / pi^(1/3))^2 ) = 256 pi^(2/3) / 16 pi = 16 / pi^(1/3)
The cylinder with minimum cost has radius 4 / pi^(1/3) = 2.73 and height 16 / pi^(1/3) = 10.92
h = 256/(pi * r^2)
The combined area of the circles is 2 * pi * r^2
The area of the side is the circumference of the circle times the height, or 2r * pi * h = 2r * pi * 256 / (pi * r^2) = 512/r
The cost per unit area of the side is some constant C; the cost per unit area of the circles is 2C.
So the total cost is 2C(2 pi r^2) + C (512/r) = 4C pi r^2 + 512C/r
The derivative of the cost is 8C pi r + -512C / r^2; set equal to zero and solve for r.
8C pi r + -512C / r^2 = 0
8C pi r = 512C / r^2
8C pi r^3 = 512C
r^3 = 64 / pi
r = 4 / pi^(1/3)
h = 256/(pi * r^2) = 256 / (pi * (4 / pi^(1/3))^2 ) = 256 pi^(2/3) / 16 pi = 16 / pi^(1/3)
The cylinder with minimum cost has radius 4 / pi^(1/3) = 2.73 and height 16 / pi^(1/3) = 10.92