1. ## Minimum Area

Need some help with this one.

$\overline {PQ}$ and $\overline {SR}$ intersect two parallel lines, and each other at a point $T$ on the interior of the parallel lines. Point $R$ is $d$ units from $P$. How far from $Q$ should the point $S$ be so that the sum of the triangles $\triangle STQ$ and $\triangle PTR$ is a minimum?

I'm having some trouble with the setup. Any help would be appreciated.
Thanks.

2. Originally Posted by McScruffy
Need some help with this one.

$\overline {PQ}$ and $\overline {SR}$ intersect two parallel lines, and each other at a point $T$ on the interior of the parallel lines. Point $R$ is $d$ units from $P$. How far from $Q$ should the point $S$ be so that the sum of the triangles $\triangle STQ$ and $\triangle PTR$ is a minimum?

I'm having some trouble with the setup. Any help would be appreciated.
Thanks.

However, given information says that:
There are two parallel lines.
P is a point on 1st parallel line, and Q on the other.
R is on the same line as P, and S on the same line as Q.
PQ and SR intersect at T.

3. That is all of the information given in the problem. I can visualize the problem, the trouble that I'm having is getting an equation to differentiate.

4. Let,

m be the distance between S and Q (which we have to eventually determine)

n be the distance between parallel lines

$h_1$ be the altitude of \triangle PTR i.e. area of \triangle PTR= $\frac{1}{2}dh_1$

$h_2$ be the altitude of \triangle STQ i.e. area of \triangle STQ= $\frac{1}{2}mh_2$

hence, sum= $\frac{1}{2}dh_1+\frac{1}{2}mh_2$...........(1)

Now,
$h_1=\frac{nd}{d+m}$

$h_2=\frac{nm}{d+m}$

Substitute value of $h_1$ and $h_2$ in equation 1 and differentiate with respect to m to maximize area.