A figure is bounded by the curves `y=x^2+1, y=0,x=0,a n dx=1.` At what point `(a , b)` should a tangent be drawn to curve `y=x^2+1` for it to cut off a trapezium of greatest area from the figure?

Answer

`(1//2,5//4)`

Explanation

Equation of the curve is y =`x^(2)+1`
Tangent at P(a,b) is y-b = 2a(x-a)
i.e `y-(a^(2)+1) is y -b =2a(x-a)`
`x=ararry=1-a^(2)` which is positive for `0ltalt1`
and `x=1 rarr y =1 + 2a-a^(2)`

Z=Area of trapezium OABC
`=1/2 (OC+AB)OA=1+a-a^(2),0ltalt1`
`(dZ)/(da)=[1-2a]=0`
or `a=1//2 and (d^(2)Z)/(da^(2))=-4lt0`
Thus at a=`1//2` area of trapezium is maximum (greatest ).Thus the requaired points is `(1//2,5//4)`