They have provided you with all the formulas you need
given: πr^2h = 128
h = 128/(πr^2)
so in
SA = 2πr^2+ 2πrh
= 2πr^2+ 2πr(128/(πr^2)
= 2πr^2 + 256/r
d(SA)/dr = 4πr - 256/r^2
= 0 for a min SA
4πr = 256/r^2
r^3 =64/π
r = 4/π^(1/3) = appr 2.7311
then h = 128/(πr^2) = 5.46223
notice that this is twice the radius
So the minimum SA is obtained when the
radius is 2.4311 cm
and the height is 5.46223 cm
Optimization Problem
A right circular cylindrical can of volume 128tπ cm^3 is to be manufactured by a company to store their newest kind of soup. They want to minimize the surface area of the can to keep costs down.
What are the dimensions of the can with minimum surface area?
The volume of a cyllinder is V= πr^2h, where r is the radius and h is height. The surface area of a cylinder is SA= 2πr^2+2πrh, which is the sum of the area of the top and bottom (2 circles) and the area of the other curved sides (a rectangle, whose length is the circumference of the circles)
2 answers
given: πr^2h = 128π
h = 128π/(πr^2)
TSA = 2πr^2+ 2πrh
= 2πr^2+ 2πr(128/(r^2))
d(tsa)/dr = 4πr - (256π/r^2)
0 = 4πr - (256π/r^2)
4πr = (256π/r^2)
(solve for r)
r^3 = 64
r = 4
(solve for h)
h = 128/(4^2)
= 8
h = 128π/(πr^2)
TSA = 2πr^2+ 2πrh
= 2πr^2+ 2πr(128/(r^2))
d(tsa)/dr = 4πr - (256π/r^2)
0 = 4πr - (256π/r^2)
4πr = (256π/r^2)
(solve for r)
r^3 = 64
r = 4
(solve for h)
h = 128/(4^2)
= 8
5 answers