Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.7 OPTIMIZATION PROBLEMS ¤ 193
and if c>0, it represents a maximum value. As |c| increases, the maximum or
minimum point gets closer to the origin. To find the inflection points, we
differentiate again: f 0 (x) =e −cx (1 − cx)
⇒
f 00 (x) =e −cx (−c)+(1− cx)(−ce −cx )=(cx − 2)ce −cx . This changes sign
when cx − 2=0 ⇔ x =2/c. Soas|c| increases, the points of inflection get
closer to the origin.
37. (a) f(x) =cx 4 − 2x 2 +1.Forc =0, f(x) =−2x 2 +1, a parabola whose vertex, (0, 1), is the absolute maximum. For
c>0, f(x) =cx 4 − 2x 2 +1opens upward with two minimum points. As c → 0, the minimum points spread apart and
move downward; they are below the x-axis for 0 0, so there is a local minimum at
x = ±1/ √
c.Heref ±1/ √
c = c(1/c 2 ) − 2/c +1=−1/c +1.
But ±1/ √
c, −1/c +1 lies on y =1− x 2 since 1 − ±1/ √ 2
c =1− 1/c.
4.7 Optimization Problems
1. (a)
First Number Second Number Product
1 22 22
2 21 42
3 20 60
4 19 76
5 18 90
6 17 102
7 16 112
8 15 120
9 14 126
10 13 130
11 12 132
We needn’t consider pairs where the first number
is larger than the second, since we can just
interchange the numbers in such cases. The
answer appears to be 11 and 12, but we have
considered only integers in the table.