Let the distance for the connection be x down the main pipe line. Therefore the distance form the connection to the power plant is (10-x).

The distance from the new well to the connection is `sqrt(3^2+x^2) = sqrt(9+x^2)`

the time required to flow is t, then

`t = (10-x)/15...

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Let the distance for the connection be x down the main pipe line. Therefore the distance form the connection to the power plant is (10-x).

The distance from the new well to the connection is `sqrt(3^2+x^2) = sqrt(9+x^2)`

the time required to flow is t, then

`t = (10-x)/15 + sqrt(9+x^2)/10`

`(dt)/(dx) = (-1)/15 + 1/10*1/2*2x(9+x^2)^(-1/2)`

`(dt)/(dx) = (-1)/15 + x/(10sqrt(9+x^2))`

For critcal points, dt/dx = 0

then,

`(-1)/15 + x/(10sqrt(9+x^2)) = 0`

`x/(10sqrt(9+x^2)) = 1/15`

`3x = 2sqrt(9+x^2)`

by taking the square of each side,

`9x^2 = 4(9+x^2)`

`5x^2 = 36` this gives `x = 6/sqrt(5)`

To check whether this is a minimum, we have to check for the sign of the second derivative, `(d^2t)/(dx^2)`

`(d^2t)/(dx^2) = 1/10[(sqrt(9+x^2)-x*1/2*2x*(9+x^2)^((-1)/2))/(9+x^2)]`

`(d^2t)/(dx^2) = 1/10[(sqrt(9+x^2)-x^2/sqrt(9+x^2))/(9+x^2)]`

`(d^2t)/(dx^2) = 1/10[((9+x^2)-x^2)/(9+x^2)^(3/2)]`

`(d^2t)/(dx^2) = 9/(10(9+x^2)^(3/2))`

This is positive for any value there for at `x =6/sqrt(5)` the second derivative is positive and we have a minimum value there for the function t.

Therefore to minimize the time the engineers have to fix the connection `6/sqrt(5)` distance down the main pipeline.