SOLUTION: You have been asked to design a can shaped like a right circular cylinder with height h and radius r. Given that the can must hold exactly 410 cm3, what values of h and r will mini

Algebra ->  Volume -> SOLUTION: You have been asked to design a can shaped like a right circular cylinder with height h and radius r. Given that the can must hold exactly 410 cm3, what values of h and r will mini      Log On


   

Question 1033164: You have been asked to design a can shaped like a right circular cylinder with height h and radius r. Given that the can must hold exactly 410 cm3, what values of h and r will minimise the total surface area (including the top and bottom faces)? Give your answers correct to 2 decimal places as a list [in brackets] of the form: [ h, r ]
for constants h (height), r (radius), in that order.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
the volume of the cylinder with height h and radius r (both in cm) is
pi%2Ar%5E2%2Ah=410 cm%5E3 .
pi%2Ar%5E2%2Ah=410 ---> h=410%2F%28pi%2Ar%5E2%29 .
The total surface area (including the top and bottom faces)of a right circular cylinder with height h and radius r is
A=2pi%2Ar%2Ah%2B2pi%2Ar%5E2 .
(We would measure h and radius r in cm, and A would be in cm%5E2, of course).
Substituting the expression found for h ,
A=2pi%2Ar%2A%28410%2F%28pi%2Ar%5E2%29%29%2B2pi%2Ar%5E2 <--> A=2%2A410%2Fr%2B2pi%2Ar%5E2 <--> A=2%2A%28410%2Bpi%2Ar%5E3%29%2Fr .

We need to find the value of r that yields the minimum for A .
There may be another way to find that value.
Maybe you are expected to do it using a graphing calculator,
or maybe you are expected to use calculus,
and specifically derivatives.
The result should be the same.
Using calculus:
A local minimum of A happens only for a value of r that makes the derivative zero.
The derivative of A=2%2A410%2Fr%2B2pi%2Ar%5E2 is
dA%2Fdr=-2%2A410%2Fr%5E2%2B4pi%2Ar=%282%2Fr%5E2%29%28-410%2B2pi%2Ar%5E3%29 .
%282%2Fr%5E2%29%28-410%2B2pi%2Ar%5E3%29=0<--->-410%2B2pi%2Ar%5E3=0<--->r%5E3=410%2F%282%2Api%29=205%2Fpi ---> r=root%283%2C%28205%2Fpi%29%29=highlight%284.03%29 (correct to 2 decimal places).
Then,