# Thread: Minimum Area

1. ## Minimum Area

Need some help with this one.

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

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

I'm having some trouble with the setup. Any help would be appreciated.
Thanks.
Do you have some more information???

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

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

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

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

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

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

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