Question

The vertices of a triangle are $(0,0),(x,\cos x)$ and $({\sin }^{3}x,0)$ where $0<x<\frac{\pi }{2}$. The maximum area for such a triangle in sq. units, is

• A. $\frac{3\sqrt{3}}{32}$
• B. $\frac{\sqrt{3}}{32}$
• C. $\frac{4}{32}$
• D. $\frac{6\sqrt{3}}{32}$

Answer 1

The correct option is A $\frac{3\sqrt{3}}{32}$

$f\left(x\right)=\frac{{\sin }^{3}x\cos x}{2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{3{\sin }^{2}x{\cos }^{2}x-{\sin }^{4}x}{2}$
$f^{\prime }\left(x\right)=0\Rightarrow 3{\sin }^{2}x{\cos }^{2}x-{\sin }^{4}x=0$
$\Rightarrow 3{\cos }^{2}x-{\sin }^{2}x=0$
$\Rightarrow 4{\cos }^{2}x-1=0$
$\Rightarrow \cos x=\frac{1}{2}$
$\Rightarrow x=\frac{\pi }{3}$, which is the point of maxima.
$\Rightarrow f_{max}=\frac{\left(\sqrt{3}/2{)}^3\right(1/2)}{2}=\frac{3\sqrt{3}}{32}$