Question

The equation of the line through (3, 4) which cuts from the first quadrant a triangle of minimum area is

• A. 4x + 3y - 24 = 0
• B. 3x + 4y - 12 = 0
• C. 2x + 3y - 12 = 0
• D. 3x + 2y - 24 = 0

Answer 1

Answer: (A)

4x + 3y - 24 = 0

Let it interscept at $(x,0)(0,y)$

Slope$=\frac{-y}{x}=\frac{4}{3-x}$

$area=\frac{1}{2}xy$

$=\frac{1}{2}x\left(\frac{4x}{x-3}\right)$

$\frac{d\left(area\right)}{dx}=2(\frac{2x}{(x-3)}-\frac{x^2}{(x-3{)}^2}$

$0=2\left(\frac{x}{x-3}\right)\left(\frac{2(x-3)-x}{(x-3)}\right)$

$0=\frac{2x(x-6)}{(x-3{)}^2}$

$x=0,6$

$\frac{d\left(area\right)}{dx}<0$ If $0<x<6$

$\frac{d\left(area\right)}{dx}>0$ If $x<0$ or $x>6$

So area is min when $x=6,y=8$

So Line $=4x+3y-24=0$