5.1. OPTIMIZATION WITH DERIVATIVES 183
We know r > 0. Looking at the sign chart for the derivative, plugging in r D 2 and r D 4, we
get:
.0/ .3:17/ C C C .1/
so the critical value x D 3:17 is a minimum. Plugging the r value into the formula for h, we get
that
h
Š
6:34
. Our answer is
r
D
3:17
cm and
h
D
6:34
cm.
˙
Let’s see how the second derivative test shakes out in the previous example.
A
00
D 4 C
800
r
3
which is positive for any r > 0. So the curve is concave up in the possible region, and our
critical value is a minimum; the second derivative test agrees with the sign chart for the first
derivative test.
When you are working optimization story problems, it is critical to make sure the values you get
make sense. Negative lengths, for example, probably mean you made a mistake. ese problems
also have the property that some of what you did may be needed again in a later step. Using a
neat layout, possibly informed by the steps given in Knowledge Box 5.5, will help.
PROBLEMS
Problem 5.9 For each of the following functions, find the global maximum and minimum of
the function, if they exist, on the stated interval.
1. f .x/ D 2x 1 on .2; 2/
2. g.x/ D 3x C 1 on Œ3; 1
3. h.x/ D x
2
C 4x C 3 on Œ1; 4
4. r.x/ D ln.x/ on Œ1; e
3
5. s.x/ D e
x
2
, on Œ2; 3
6. q.x/ D x
3
16x C 1, on Œ4; 4