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How to find the maximum value of $12\sin x -9\sin^2x$ ;

My approach :

This can be written as $-[(3\sin x -2)^2-4]$.
It means that the function will be maximum when $(3\sin x-2)^2 <4$ due to negative sign outside bracket.

But I am not getting how to proceed from here, please suggest. Thanks.

Vincenzo Tibullo
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user108258
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3 Answers3

4

Note $$ 12\sin x-9\sin^2x=3\sin x(4-3\sin x)\le \left(\frac{3\sin x+(4-3\sin x)}{2}\right)^2=4 $$ and the equal sign holds if and only if $3\sin x=4-3\sin x$ or $\sin x=\frac{2}{3}$. Thus the max is 4 when $\sin x=\frac{2}{3}$.

hola
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xpaul
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  • Nice solution Xpaul Using Inequality, But i did not understand bcz $\bf{A.M\geq G.M}$ is applicable when $3\sin ,4-3\sin x>0$ but here $3\sin x $ may be positive or negative, bcz $x$ not given here. – juantheron Feb 20 '16 at 06:12
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Hint The function $f(x) = 12\sin(x) - 9\sin^2(x)$ has a local maximum wherever

$$0 = f'(x) = 12\cos(x) - 18\sin(x)\cos(x)$$

which is true for $\cos(x) = 0$ or $\sin(x) = \frac{12}{18}$

Since the function is periodic you need to check a few local max points, (but not too many).

naslundx
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1

You can start by making a perfect square

-9$(sin^2x$-$\frac{4}{3}sinx$)

$\implies$-9$((sinx+\frac{2}{3})^2-(\frac{2}{3})^2)$

$\implies$ 4 - 9$([-1,1]+\frac{2}{3})^2$

$\implies$ 4 - $[0,\frac{5}{9}]$

$\implies$ $[\frac{31}{9},4]$

Therefore, Maximum value=4

Abhishekstudent
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