Page 153 - How To Solve Word Problems In Calculus

dθ
Setting = 0,
dx

2 − 6x 2
0 =
2 2
(1 + 3x )
0 = 2 − 6x 2

2
6x = 2
√
1 3
x = √ =
3 3

√
dθ 3
2
Since sec θ> 0, is positive if x < and negative if
√ dx 3
3
x > . By the first derivative test, θ has a relative maximum
√ 3
at 3/3. Since it is the only relative extremum on the interval
[0, ∞), it must give the absolute maximum.
√
2x 3
Since tan θ = , the value x = gives
1 + 3x 2 3

√
2 3
√
3 3
tan θ = =
1 3
1 + 3 ·
3

The corresponding value of θ is π/6. The maximum angle of
sight is 30 .
◦

EXAMPLE 8
Find the maximum area of a rectangle circumscribed about a
fixed rectangle of length 8 and width 4.

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