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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?

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I just need help with setting up the function.

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  • $\begingroup$ 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 $\endgroup$ – Trash Darryl Feb 28 at 10:39
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    $\begingroup$ This is a classic (so it must a duplicate). Hint: reflect $B$ across $MN$, in a manner similar to math.stackexchange.com/questions/3601626/…. $\endgroup$ – player3236 Feb 28 at 10:39
  • $\begingroup$ Do you have to use calculus? $\endgroup$ – Math Lover Feb 28 at 10:40
  • $\begingroup$ Yeh the question has to be solved using calculus $\endgroup$ – Trash Darryl Feb 28 at 10:44
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    $\begingroup$ 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. $\endgroup$ – Math Lover Feb 28 at 10:48
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$AM+XB$ 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

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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

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