# 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.

## 1 Answer

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 $k=2$ 