find-the-minimum-and-maximum-value-of-5-f-3-where-f-8cos-15-sin-

Question Number 153870 by physicstutes last updated on 11/Sep/21

find the minimum and maximum value

of \frac{5}{f (θ) + 3} where f (θ) = 8 \cos θ - 15 \sin θ

Commented by mr W last updated on 12/Sep/21

t h e r e a r e n o m i n i m u m a n d n o

m a x i m u m . t h e r e a r e o n l y l o c a l

m i n i m u m a n d l o c a l m a x i m u m .

Commented by physicstutes last updated on 12/Sep/21

why sir?

Commented by mr W last updated on 12/Sep/21

f (θ) = 8 \cos θ - 15 \sin θ = 17 \cos (θ + α)

- 17 ⩽ f (θ) ⩽ 17

t h a t m e a n s f (θ) c a n b e - 3,

w h e n f (θ) \to - 3^{-}, \frac{5}{f (θ) + 3} \to - \infty \Rightarrow n o m i n i m u m

w h e n f (θ) \to - 3^{+}, \frac{5}{f (θ) + 3} \to + \infty \Rightarrow n o m a x i m u m

Answered by liberty last updated on 11/Sep/21

g (θ) = \frac{5}{f (θ) + 3} = \frac{5}{17 \cos (θ - ϑ) + 3}

\Rightarrow 17 \cos (θ - ϑ) + 3 = \frac{5}{g (θ)}

\Rightarrow 17 \cos (θ - ϑ) = \frac{5}{g (θ)} - 3

w e k n o w t h a t - 17 ⩽ 17 \cos (θ - ϑ) ⩽ 17

- 17 ⩽ \frac{5}{g (θ)} - 3 ⩽ 17

- 14 ⩽ \frac{5}{g (θ)} ⩽ 20

- \frac{14}{5} ⩽ \frac{1}{g (θ)} ⩽ 4

\frac{1}{4} ⩽ g (θ) < \infty

m i n i m u m = \frac{1}{4}

h a s n o m a x i m u m

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