Let R = refinery, S = storage tanks, Q = point on south bank opposite P.
Let RP = x, then, QS = 3 - x, and we already know PQ = 1.
Using Pythagoras in triangle PQS, PS^2 = 1^2 + (3 - x)^2, so PS = âš(x^2 - 6x + 10).
Total distance of pipeline is therefore: D = x + âš(x^2 - 6x + 10).
Total cost is thus: C = 200000x + 400000âš(x^2 - 6x + 10).
To minimise this, take the derivative and set it equal to zero.
dC/dx = 200000 + 400000(1/2)[1/âš(x^2 - 6x + 10)](2x - 6) = 0
Simplifying gives: 3x^2 - 18x + 26 = 0
Solving gives: x = (9 ± âš3)/3
Discard the positively signed solution as this gives x > 3 which is invalid.
Therefore, x = (9 - âš3)/3 ≠2.42 km
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