Question:

Two quadratics both pass through opposite corners of a rectangle whose sides are parallel to the xy-Cartesian axes. One quadratic has it's vertex on the lower left corner, but also passes through the upper right corner, and the second has it's vertex on the upper right, but also passes through the lower left. These two quadratics form a region within the rectangle. What fraction of the area of the rectangle is the area of this region?

Answer:Relocate the coordinate axes so that the origin lies on the lower left hand corner. The upper right hand corner is at a point `(m,n)`. The equation of one polynomial is `(n/m^2)x^2`, the equation of the other is `(n/m^2)(2mx-x^2)`. This results in the following integral: `int_0^m (n/m^2)(2mx-2x^2) dx = (2n/m^2)int_0^m mx - x^2 dx` evaluating gives `(2n/m^2)(m^3/2 - m^3/3)=1/3 mn ` ,so the total fraction of the area is `1/3`.

## No comments:

## Post a Comment