Find the smallest value of f(x) := (\frac19+\frac{32}{\sin(x)})(\frac{1}{32}+\frac{9}{\cos x}) on

iocasq4

iocasq4

Answered question

2022-01-23

Find the smallest value of f(x)=(19+32sin(x))(132+9cosx) on the interval (0,π2)

Answer & Explanation

sphwngzt

sphwngzt

Beginner2022-01-24Added 11 answers

Since both cos and sin are positive in (0,π2) we can use Cauchy inequaliy:
(a2+b2)(c2+d2)(ac+bd)2
(19+32sin(x))(132+9cosx)(1288+288sinxcosx)2(1122+24)2
We used here
sin(x)cos(x)=12sin(2x)12
with equality at x=π4. So
ymin=(1122+24)2

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