Integration Area - Parabola and a line Questions (1 Viewer)

[2013_jonathan_s]

The points A(3,9) and B(-2,4) lie on the parabola y=x^2. The line y=x+6 joins A and B. The point P(p,p^2) is a variable point on the parabola below the line. Show that the greatest possible area of the triangle APB is three-quarters of the area of the parabolic segment APB, given that the area of parabolic segment is integral from -2 to 3 of x^2 dx.

Thanks.

 

[pdang]

For the triangle APB, the base length is distance AB and the height you get using the perp dist eqn using P and x-y+6=0.
Using area=bh/2, you can get an equation for the area in terms of p, which you can differentiate and find the max value of.
Evaluate the given integral and show that the max area you got before is 3/4 of this.