The maximum value of 12 sin theta - 9 sin^2 t | vedclass

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Question

(B) Let $f(	heta) = 12 \sin 	heta - 9 \sin^2 	heta$.To find the maximum value,let $x = \sin 	heta$,where $x \in [-1, 1]$.Then $f(x) = 12x - 9x^2$.

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $f(	heta) = 12 \sin 	heta - 9 \sin^2 	heta$.To find the maximum value,let $x = \sin 	heta$,where $x \in [-1, 1]$.Then $f(x) = 12x - 9x^2$.To find the critical points,take the derivative with respect to $x$:$f'(x) = 12 - 18x$.Setting $f'(x) = 0$ gives $12 - 18x = 0$,so $x = \frac{12}{18} = \frac{2}{3}$.Since $\frac{2}{3} \in [-1, 1]$,we evaluate the function at this point:$f\left(\frac{2}{3}ight) = 12\left(\frac{2}{3}ight) - 9\left(\frac{2}{3}ight)^2 = 8 - 9\left(\frac{4}{9}ight) = 8 - 4 = 4$.Since $f''(x) = -18 Thus,the maximum value is $4$.

The maximum value of $12 \sin \theta - 9 \sin^2 \theta$ is

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