Cal Prob~

At 4:00pm the second boat is at 15km due west of the dock.
At x minutes after 4:00pm, the 2nd boat is at 15 - 15*(x/60) or 15 - x/4 due west of the dock.
At that same time, the first boat is at 20*(x/60) or x/3 due south of the dock
Their distance at time x is D where D^2 = S = (15 - x/4)^2 + (x/3)^2
dS/dx = 2(15 - x/4)(-1/4) + (2x/3)(1/3) = x/8 - 15/2 + 2x/9 = 25x/72 - 15/2
For min, dS/dx = 0 => 25x/72 - 15/2 = 0
25x/72 = 15/2 => x = 21.6
d^S/dx^2 = 25/72 > 0 => minimm
The closest distance happens at around 22 minutes after 4:00pm
Let P be located at x km away from the refinery in the north bank.
The cost of the pipe in the north bank = 400000x
The length of pipe under the river = √[(7 - x)^2 + 2^2] = √(x^2 - 14x + 53)
The cost of the pipe under the river = 800000√(x^2 - 14x + 53)
Total cost C = 400000x + 800000√(x^2 - 14x + 53)
C = 400000[x + 2√(x^2 - 14x + 53)]
dC/dx = 400000[1 + (2x - 14)/√(x^2 - 14x + 53)] = 0
1 = (14 - 2x)/√(x^2 - 14x + 53)
√(x^2 - 14x + 53) = 14 - 2x
x^2 - 14x + 53 = (14 - 2x)^2
x^2 - 14x + 53 = 196 - 56x + 4x^2
3x^2 - 42x + 143 = 0
x = {42 +/- √[42^2 - 4(3)(143)]}/6
x = 7 +/- (√48) / 6
x = 8.15 (rejected) or x = 5.85
When x = 5.85, C = 4185643
when x = 5.86, C = 4185668
when x = 5.84, C = 4185644
Therefore the minimuj occurs when P is 5.85 km away from the refinery