# A right triangle in 1st quad has its base on x-axis & is formed by joining (0,0) to a point on the parabola y=4-x^2 and dropping a perpendicular on the x-axis. What is the maximum area of the...

A right triangle in 1st quad has its base on x-axis & is formed by joining (0,0) to a point on the parabola y=4-x^2 and dropping a perpendicular on the x-axis. What is the maximum area of the triangle.

*print*Print*list*Cite

### 1 Answer

A triangle is formed in the first quadrant by joining the points (0, 0), (X, Y) and (X, 0) with (X, Y) lying on the parabola y = 4 - x^2.

The area of the triangle is `A = (1/2)*b*h = (1/2)*X*Y = (1/2)*X*(4 - X^2)`

To maximize the area solve `(dA)/(dX) = 0`

=> `(1/2)*(4 - 3X^2) = 0`

=> `X^2 = 4/3`

=> `X = 2/sqrt 3`

Y = `4 - 4/3 = 8/3`

The area of the triangle is `(1/2)*(2/sqrt 3)*(8/3) = (8*sqrt 3)/9`

**The maximum area of the triangle is **`(8*sqrt 3)/9`