Point A lies on the curve y = e^{-x^2} and ha | vedclass

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(D) The vertices of the triangle $AOB$ are $O(0, 0)$,$B(x, 0)$,and $A(x, e^{-x^2})$.Since the triangle is right-angled at $B$,the area $A(x)$ is given

Vedclass · Accepted Answer

$\frac{1}{\sqrt{8e}}$

Vedclass · Answer

(D) The vertices of the triangle $AOB$ are $O(0, 0)$,$B(x, 0)$,and $A(x, e^{-x^2})$.Since the triangle is right-angled at $B$,the area $A(x)$ is given by:$A(x) = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes x 	imes e^{-x^2} = \frac{x e^{-x^2}}{2}$.To find the maximum area,we differentiate $A(x)$ with respect to $x$:$A'(x) = \frac{1}{2} [1 \cdot e^{-x^2} + x \cdot e^{-x^2} \cdot (-2x)] = \frac{e^{-x^2}}{2} [1 - 2x^2]$.Setting $A'(x) = 0$ for critical points:$1 - 2x^2 = 0 \implies x^2 = \frac{1}{2} \implies x = \frac{1}{\sqrt{2}}$ (since $x > 0$).Now,calculate the maximum area at $x = \frac{1}{\sqrt{2}}$:$A_{max} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \cdot e^{-(1/\sqrt{2})^2} = \frac{1}{2\sqrt{2}} \cdot e^{-1/2} = \frac{1}{2\sqrt{2} \sqrt{e}} = \frac{1}{\sqrt{8e}}$.Thus,the maximum area is $\frac{1}{\sqrt{8e}}$.

Point $A$ lies on the curve $y = e^{-x^2}$ and has the coordinates $(x, e^{-x^2})$ where $x > 0$. Point $B$ has the coordinates $(x, 0)$. If $O$ is the origin,then the maximum area of the triangle $AOB$ is

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