The tangent to the parabola `y=x^2` has been drawn so that the abscissa `x_0` of the point of tangency belongs to the interval [1,2]. Find `x_0` for which the triangle bounded by the tangent, the axis of ordinates, and the straight line `y=x0 2` has the greatest area.

Explanation

`y=x^(2)` or `dy//dx=2x`
Therefore , 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 is `(0,x_(0)^(2))`. Thus
Z= area of triangle PQR
`=(1)/(2)2x_(0)^(2)x_(0)=x_(0)^(3),1lex_(0)le2`
`therefore (xz)/(dx_(0))=3x_(0)^(2)gt0` in `1 le x_(0) le le2`
Thus , Z is an increasing function in [1,2]
Hence Z, i.e the area of `trianglePQR`, is greatest at `x_(0)=2`.