Posted by **Santha** on Tuesday, March 16, 2010 at 2:30pm.

Find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area.

- calculus HELP URGENT -
**Reiny**, Tuesday, March 16, 2010 at 4:44pm
Draw a line through (3,4) cutting the x-axis at (x,0) and the y-axis at (0,y)

Area = (1/2(x)(y)

using slopes,

(y-4)/-3 = +4/(x-3)

xy - 3y - 4x + 12 = 12

y(x-3) = 4x

y = 4x/(x-3)

then

A = (1/2)x(4x)/x-3)

= 2x^2/(x-3)

dA/dx = [(x-3)(4x) - 2x^2]/(x-3)^2 = 0 for a max/min of A

4x^2 - 12x - 2x^2 = 0

2x^2 - 12x = 0

2x(x-6)= 0

x = 0 or x = 6, and y = 24/3 = 8

using (0,8) and (3,4)

slope = (8-4)/-3

= -4/3

line has equation

y = (-4/3)x + 8

slope of line =

- calculus HELP URGENT -
**Santha**, Tuesday, March 16, 2010 at 4:59pm
" using slopes,

(y-4)/-3 = +4/(x-3)

xy - 3y - 4x + 12 = 12 "

can you explain that part

- calculus HELP URGENT -
**Santha**, Tuesday, March 16, 2010 at 5:48pm
nvr mind

Anyways thanks Reiny

- calculus HELP URGENT -
**Nkuche**, Sunday, April 18, 2010 at 7:24am
Find the gradient of a line joining R(4,8) & S(5,-2).

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