posted by Sribava on .
The base of a triangle is on the x-axis, one side lies along the line y=3x, and the third side passes through the point (1,1). What is the slope of the third side if the area of the triangle is to be a minimum.
Interesting problem. Check to make sure I get this right.
So, we have a triangle with base k, and vertex where the line through (1,1) and (0,k) intersects y=3x.
The line through (1,1) and (0,k) is
(y-1)/(x-1) = -1/(k-1)
y = (x-1)/(1-k) + 1
So, that line intersects y=3x when
3x = (x-1)/(1-k) + 1
x = k/(3k-2)
y = 3k/(3k-2)
The area a of the triangle is thus
a = 1/2 k * 3k/(3k-2) = 3k^2/(6k-4)
da/dk = (6k(6k-4) - 3k^2 * 6)/(6k-4)^2
da/dk = 0 when k=0 or k = 4/3
Thus, the vertex of the triangle is where the line through (1,1) and (4/3,0) intersects y=3x
(y-1)/(x-1) = 1/(-1/3)
y = -3x + 4
That line has slope -3
. . .
-3x + 4 = 3x
6x = 4
x = 2/3
So, the triangle has base k=4/3, height h=2, area a=4/3