Question

What is the maximum value of $16\sin \theta -12{\sin }^2\theta$?

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{16}{3}$
• D. $4$

Answer 1

Answer: (C)

$\frac{16}{3}$

Given

$16\sin \theta -12{\sin }^2\theta$

Let $x=\sin \theta$

$16x-12x^2$

$4x(4-3x)$

Max function is found at $f^{\prime }\left(x\right)=0$

$\frac{d}{dx}\left(4x\right(4-3x\left)\right)=0$

$4(4-3x)+4x(-3)=0$

$16-24x=0$

$x=\frac{2}{3}$

$\therefore \sin \theta =\frac{2}{3}$

$16\left(\frac{2}{3}\right)-12\left(\frac{4}{9}\right)$

$16(\frac{2}{3}-\frac{1}{3})$

$\frac{16}{3}$