# Find area of triangle which sides is limited by two functions and the x axis

I'm studying for my math exam and I'm stuck on the following question

"A triangle is limited by the x axis and the two functions $y=kx$ och $y=\frac{1}{k}x+k$ where k > 1. Determine the smalest possible area of the triangle "

I found that the functions intersect at $x=\frac{k^2}{k^2-1}$ and $y=\frac{k^3}{k^2-1}$

I'm now thinking that the height of the triangle is the y value of the point of intersection. But I don't know how to get the base of the triangle.

I looked at the answer key and it says " the side of the triangle which is on the x axis has the lenght $k^2$and the height $y=\frac{k^3}{k^2-1}$"

I don't understand how they arrive at this equation for the base of the triangle.

Note that the length of the base is the length of the side that is coincide with the x axis, and to find this length enough to examine the points where these two functions intersect the x axis.

The first fucntion clearly intersects the x as at $x=0$, however the secons intersects the x axis at $x=-k^2$. Hence the length of the base is $k^2$.

I think this plot will clarify the idea: taking $k=2$