# Thread: Calculating Triangle Area Variable

1. ## Calculating Triangle Area Variable

Bit of trouble differentiating this one. Can anyone help? Many thanks.

Q.
A curve has equation y = x2. Three points form the triangle apb. p(x,y) is a point on the curve, a, on the x-axis, is (6,0) & a final point b, also on the x-axis. bp $\perp$ ab. (i) Express the coordinates of p, in terms of x only, (ii) Find the value of x if the area of the triangle abp is a maximum and hence find this maximum area.

Attempt: (i) If y = x2, then p = (x, x2)
(ii) If bp $\perp$ ab, we can infer that b = (x,0).
Area of a triangle apb on a plane is: $\frac{1}{2}$[(y2 - y1)(x1 - x3) - (x2 - x1)(y1 - y3)] => $\frac{1}{2}$[(x2 - 0)(6 - x) - (x - 6)(0 - 0)] => $\frac{1}{2}$[6x2 - x3]
$\frac{dA}{dx}$ = $\frac{1}{2}$(12x - 3x2) = 0 => 12x - 3x2 = 0 => 4x - x2 = 0 => x2 = 4x => x = 4

Ans: (From text book): x = 2

3. ## Re: Calculating Triangle Area Variable

Apologies. I have re-edited post accordingly.

4. ## Re: Calculating Triangle Area Variable

If B is allowed to be located right of A, then there is no maximum triangle area. If B has to be left of A, then I agree that the maximum is reached when x = 4. When x = 2, the area is 8, but when x = 4, the area is 16.

Thank you.