Question

The tangent to the parabola $y=x^2$ has been drawn so that the abscissa $x_0$ of the point of tangency belong to the interval $[1,2].$ The $x_0$ for which the triangle bounded by the tangent, the axis of ordinates and the straight line $y=x_0^2$ has the greatest area is

Answer 1

$y=x^2\Rightarrow \frac{dy}{dx}=2x$
$\Rightarrow$Equation of the tangent at $(x_0,x_0^2)$ is $y-x_0^2=2x_0(x-x_0).$
It meets $y-$axis in $R(0,-x_0^2),Q\text{is}(0,x_0^2)$
$\Rightarrow Z=$ area of the triangle $PQR$
$=\frac{1}{2}2x_0^2{x}_0=x_0^3,1\le x_0\le 2$
$\frac{dZ}{d{x}_0}=3x_0^2$ in $1\le x_0\le 2$
$\Rightarrow Z$ is an increasing function in $[1,2].$
Hence, $Z,$ i.e., the area of $\Delta PQR$ is greatest at $x_0=2.$