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Q.
The coordinates of the point P(x, y) of $y = e^{-|x|}$ so that the area formed by the coordinate axes and the tangent at P is greatest, are
Application of Integrals
Solution:
Grayph of $y = e^{-|x|}$ is symmetrical about y-axis, so we coxnsider $x \ge 0$, then $y = e^{-x} \Rightarrow \frac{dy}{dx} = -e^{x}$
Tangent at P is $Y - y = -e^{-x} \left(X-x\right)$
Its x-intercept $= x + ye^{x}$ and y-intercept $= y + xe^{-x}$
So area $A = \frac{1}{2}\left(x+ye^{x}\right)\left(y+xe^{-x}\right)$
$= \frac{1}{2}\left(x +1\right)\left(1 + x\right)e^{-x}\quad\left[\because y = e^{-x}\right]$
$\frac{1}{2}\left(1 + x\right)^{2} e^{-x}$
$\frac{dA}{dx} = \left(1 + x\right)e^{-x} - \frac{1}{2}\left(1 + x\right)^{2}e^{-x} = \frac{1}{2}\left(x +1\right)\left(1 -x\right)e^{-x}$
where A is maximum if $x = 1$
So P is $\left(1, \,e^{-1}\right)$ Due to symmetry, there is another point $\left(-1,\, e^{-1}\right)$
