Page 137 - How To Solve Word Problems In Calculus

15.

r
r

h

r

The surface area of the can consists of the lateral surface area and
the area of the top and bottom circles.

S = 2πrh + 2πr 2

lateral area of
surface top and
area bottom
3
Because the volume of the can must be 1000 in ,
2
πr h = 1000
1000
h =
πr 2
Substituting into the equation for S, we obtain a function of r .

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

We differentiate with respect to r and find the critical values by
setting the derivative to 0.

S (r ) =−2000r −2 + 4πr
−2000
0 = + 4πr
r 2
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