Page 139 - How To Solve Word Problems In Calculus

2
used in the construction of the top or bottom is (2r ) . (We must
consider the wasted material as well as the material used for the
circular end.)
Hence
2
A = 2πrh + (2r ) + (2r ) 2
= 2πrh + 8r 2
3
2
Since the volume of the can is to be 1000 in , πr h = 1000 and
1000
h = . It follows that
πr 2

1000
A(r ) = 2πr + 8r 2
πr 2
2000
= + 8r 2
r
= 2000r −1 + 8r 2 0 < r < ∞

Differentiating and finding the critical number,
A (r ) =−2000r −2 + 16r

−2000
0 = + 16r
r 2

2000
= 16r
r 2
3
16r = 2000
3
r = 125
r = 5

1000 1000 40
h = = = in
πr 2 25π π
We use the second derivative test to confirm our suspicions that
this is the radius that requires the minimum amount of material
(optional).

A (r ) = 4000r −3 + 16
4000
= + 16
r 3
126