Page 162 - How To Solve Word Problems In Calculus

2
dθ (x + 4)3 − 3x(2x)
2
sec θ =
2
dx (x + 4) 2
12 − 3x 2
=
2
(x + 4) 2
dθ
To find the critical value, we set = 0
dx
12 − 3x 2
0 =
(x + 4) 2
2
0 = 12 − 3x 2

2
3x = 12
2
x = 4
x = 2

dθ dθ
2
Since sec x > 0, > 0if0 < x < 2 and < 0if x > 2. By the
dx dx
first derivative test, x = 2 is the location of a relative maximum.
Since it is the only relative extremum on the interval (0, ∞), x = 2
is the location of the absolute maximum. Lindsay should stand 2 ft
from the wall.

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