# Thread: Calculating Variable ffrom Derivative of a Triangle

1. ## Calculating Variable ffrom Derivative of a Triangle

(i) is already solved for & answers are provided separately.

I'm using a method for differentiation from a text book that should yield root values of -1/2 & 1/2. However, my final solution for slope m yields just 1/2...? Can anyone help me solve for -1/2 using this method? Many thanks.

Q. A line L, with negative slope m, passes through the point (4,2). L interacts the x-axis at a, & the y-axis at b. (Please see attachment) (i) Find the equation of L &, hence, express the coordinates of a & b in terms of m, (ii) Find the value of m, if the area of the triangle aob is a maximum &, hence, find this area.
(i) Equation of L: mx - y - 4m + 2 = 0; a = ($\frac{4m - 2}{m}$, 0), b = (0, -4m + 2)

Attempt: (ii) Area of Triangle: $\frac{1}{2}$[(y2 - y1)(x1 - x3) - (x2 - x1)(y1 - y3)] => $\frac{1}{2}$[(0 - 0)($\frac{4m - 2}{m}$ - 0) - (0 - ($\frac{4m - 2}{m}$))(0 - (-4m + 2))] => $\frac{1}{2}$[0 - ($\frac{-4m + 2}{m}$)(4m - 2)] => $\frac{1}{2}$[$\frac{-16m^2 + 8m + 8m - 4}{m}$] => $\frac{-16m^2 + 16m - 4}{2m}$ => $\frac{-8m^2 + 8m - 2}{m}$ => $\frac{-4m^2 + 4m - 1}{m}$
$\frac{dA}{dx}$ = $\frac{8m - 4}{m^2}$ = 0 => 8m = 4 => m = $\frac{1}{2}$

Ans. (From text book): m = -$\frac{1}{2}$

2. ## Re: Calculating Variable ffrom Derivative of a Triangle

Who the heck is John Forkosh?

3. ## Re: Calculating Variable ffrom Derivative of a Triangle

Damn. I've posted the problem from a separate page that uses Latex. I thought it would carry over. I'll have to write it out again. Apologies.