Draw a diagram. Let x be the distance east to point P
on land:
d^2 = x^2 + 1^2
underwater
d^2 = 3^2 + (8-x)^2
cost is thus (in half-millions)
c = sqrt(x^2 + 1) + 2sqrt(9 + (8-x)^2)
dc/dx = x/sqrt(x^2+1) - 2(8-x)/sqrt(9+(8-x)^2)
to find where dc/dx = 0 we put all over a common denominator, and set the numerator to zero:
xsqrt(9+(8-x)^2) - 2(8-x)sqrt(x^2+1) = 0
x^2(9 + (8-x)^2) = 4(8-x)^2(x^2+1)
This is satisfied when x is about 6.3
So, P should be 6.3 km downriver from the refinery. Cost will be about $M26.55
Better check my math for the details.
An oil refinery is located 1 km north of the north bank of a straight river that is 3 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 8 km east of the refinery. The cost of laying pipe is $500,000/km over land to a point P on the north bank and $1,000,000/km under the river to the tanks. To minimize the cost of the pipeline, how far downriver from the refinery should the point P be located?
1 answer
1 answer