Minimizing Area of 2 Triangles Problem

[DMOC]

Homework Statement

The line joining P and Q crosses two parallel lines that are 8 units apart (thus, if I drew a vertical line from the top line to the bottom line, that would equal eight). The point R is 10 units from point P (as shown). How far from point Q should the point S be chosen so that the sum of the areas of the two triangles is at a minimum?

Homework Equations

Area = \frac{1}{2}bh

The Attempt at a Solution

Okay ... not so bad, I think I know some of the steps.

Area of 2 Triangles = \frac{1}{2}bh + \frac{1}{2}(b_{1})(h_{1})

b1 and h1 are just the base/height of the other trianglw (the one with points S and Q).

So should I substitute 10 in for b and then do implicit differentiation? And how do I include the "8" that's here? Have something like height x and height 8-x for the two triangles?

 

[lanedance]

you need to find the areas of the triangles in terms of the distance QS = s, before you can differentiate with respect to s to find your minima

are you sure this is everything in the question 7 how its was given? unless it comes out as a really nice result (unlikely), I think you need to know the position of Q relative to R & P first

 

[DMOC]

Yes, I coped it directly from my book.

So when doing this problem, how can I first find the areas in terms of QS? Should I just assign a variable to QS?

 

[lanedance]

yeah, so find a way to write the area in terms of s = QS,

expanding qhat you previously wrote, including the dependence on s
A_{total}(s) = A_1(s) + A_2(s) = \frac{1}{2}b_1(s)h_1(s) + \frac{1}{2}b_{2}h_{2}(s)
only the base PR = b₂, is independent of s

however unless I'm missing something I still can't see how you can do this without something pinning the location of Q

 

[Office_Shredder]

I think you have enough. Notice the two triangles will be similar

 

[lanedance]

knew i was missing something easy - nice, (can't believe i missed it)
so DMOC, its all about ratios of length