Question

a can in the shape of right circular that will hold 500cm3 of liquid.the material for the top and bottom cost 0.02$/cm2,and the sides cost 0.01$/cm2.
suppose that the tops and bottoms radius r are punched out from from square sheets with sides of length 2r and the scraps are waste.estimate the radius,hight,and cost of the can,and determine whether the can of least cost to be taller

please help

Answer 1

\[V = h*\pi r^2 = 500\]
\[cost = .02(4r^2 +4r^2) + .01(h*2 \pi r)\]
\[cost = .16 r^2 + .02 h \pi r\]
from volume equation we know that
\[\rightarrow h \pi = \frac{500}{r^2}\]
by substitution:
\[cost = .16 r^2 + \frac{10}{r}\]
minimize cost by setting derivative equal to 0
\[\rightarrow .32r - \frac{10}{r^2} = 0\]
\[\rightarrow .32 r^3 = 10\]
\[r = \sqrt[3]{\frac{10}{.32}} = \frac{5 \sqrt[3]{2}}{2}\]
you can plug that into volume equation to determine height of can

Answer 2

\[cost=.02*(4pir ^{2}+4pir ^{2})+.01*2pirh?\]

sorry,i can't type symbol pi

Answer 3

and how can i compare with the original high to realize that the can is taller?