# Minizing the total lenght of 2 pipelines, Optimization question [duplicate]

A pump house is to be placed at some point $$X$$ along a river, A pipe from point $$A$$ and a pipe from point $$B$$ will then be connected to the point $$X$$. How far should $$X$$ be away from $$M$$, so that the total length of the pipes $$\overline{AX}$$ and $$\overline{BX}$$ are minimised?

I just need help with setting up the function.

• The base of both triangle will sum up to a total of 5km and they have heights of 1km and 2km. Based on that I think I should be using the surface area formula to set up a function – Trash Darryl Feb 28 at 10:39
• This is a classic (so it must a duplicate). Hint: reflect $B$ across $MN$, in a manner similar to math.stackexchange.com/questions/3601626/…. – player3236 Feb 28 at 10:39
• Do you have to use calculus? – Math Lover Feb 28 at 10:40
• Yeh the question has to be solved using calculus – Trash Darryl Feb 28 at 10:44
• OK so if it has to be done using calculus, where are you stuck? Use Pythagoras to write $AX$ and $XB$, differentiate their sum and equate to zero. – Math Lover Feb 28 at 10:48

$$AM+XB$$ is minimum when reflexing $$B$$ in $$D$$ wrt $$MB$$ we connect $$AD$$. The point $$X$$ where $$AD$$ intersects $$MN$$ is the minimum required. Because any other point $$P$$ is such that $$AP+PD>AD$$ for the triangular inequality.

$$\triangle ACD$$ similar $$\triangle AMX$$

$$MX:AM=CD:AC$$

$$MX=\frac{AM\cdot CD}{AC}=\frac{10}{3}$$km

Here's what I did.

Hyputenuse of triangle AMX= sqrtroot of 4+×^2

Hyputenuse of triangle BNX= sqrtroot of 1+(5-x)^2

Sqrt root 4+x^2 + sqrtroot 1+(5-x)

I differentiate that and got (×/sqrroot 4+x^2) - (-x+5/sqrtroot x^2 - 10x +26)

I graphed the function and got 3.333 I then made a sign diagram which clarified that It was a local minimum