The vertices of a triangle are (0, 0),(x, cos | vedclass

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(A) The area $A$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(

Vedclass · Accepted Answer

$\frac{3\sqrt{3}}{32}$

Vedclass · Answer

(A) The area $A$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Substituting the given vertices $(0, 0), (x, \cos x), (\sin^3 x, 0)$,we get: $A(x) = \frac{1}{2} |0(\cos x - 0) + x(0 - 0) + \sin^3 x(0 - \cos x)| = \frac{1}{2} |-\sin^3 x \cos x| = \frac{1}{2} \sin^3 x \cos x$. To find the maximum area,we differentiate $A(x)$ with respect to $x$: $A'(x) = \frac{1}{2} [3 \sin^2 x \cos x \cdot \cos x + \sin^3 x(-\sin x)] = \frac{1}{2} [3 \sin^2 x \cos^2 x - \sin^4 x]$. Setting $A'(x) = 0$,we get $3 \sin^2 x \cos^2 x = \sin^4 x$. Since $0 Substituting $x = \frac{\pi}{3}$ into $A(x)$: $A(\frac{\pi}{3}) = \frac{1}{2} \sin^3(\frac{\pi}{3}) \cos(\frac{\pi}{3}) = \frac{1}{2} (\frac{\sqrt{3}}{2})^3 (\frac{1}{2}) = \frac{1}{2} \cdot \frac{3\sqrt{3}}{8} \cdot \frac{1}{2} = \frac{3\sqrt{3}}{32}$.

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:

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