Question 1142710
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Let x be the distance from the refinery to the point P on the north bank of the river.  Then the distance from point P to the storage tanks is the hypotenuse of a right triangle with legs 2 and sqrt(6-x).  Then the cost function is<br>
{{{C = 400000x + 800000*sqrt((6-x)^2+2^2)}}}
{{{C = 400000x + 800000*(x^2-12x+40)^(1/2)}}}<br>
Find the minimum cost by finding where the derivative is zero.<br>
{{{dC/dx = 400000 + 800000((1/2)(x^2-12x+40)^(-1/2)(2x-12))}}}
{{{dC/dx = 400000 + 800000((x-6)/sqrt((x^2-12x+40)))}}}<br>
Setting the derivative equal to 0 and solving...<br>
{{{400000 + 800000((x-6)/sqrt((x^2-12x+40))) = 0}}}
{{{(x-6)/sqrt((x^2-12x+40)) = -1/2}}}
{{{(x^2-12x+36)/(x^2-12x+40) = 1/4}}}
{{{4x^2-48x+144) = x^2-12x+40}}}
{{{3x^2-36x+104 = 0}}}
{{{x = (36 - sqrt(1296-1248))/6 = (36-sqrt(48))/6 = (36-4*sqrt(3))/6}}}<br>
To four decimal places that is 4.8453.<br>
ANSWER: Point P should be located about 4.8453km east of the refinery to minimize the cost of the pipeline.