Answer to Question #271118 in Calculus for Sara

Question

A cable is to be run from a power plant on one side of a river 900 meters wide to a factory

on the other side, 300 meters downstream. The cost of running the cable under the water is $5

per meter, while the cost over land is $4 per meter. What is the most economical route over

which to run the cable?

Answer 1

Let $d=$ the length of cable run under the water, $300-x=$ the length of cable run over the water.

Then

$$Cost=C(x)=5\sqrt{(900)^2+x^2}+4(300-x), 0\leq x\leq300$$

Differentiate with respect to $x$

$$C'(x)=\dfrac{5(2x)}{2\sqrt{(900)^2+x^2}}-4$$

Find the ctitical number(s)

$$C'(x)=0=>\dfrac{5(2x)}{2\sqrt{(900)^2+x^2}}-4=0$$

$$5x=4\sqrt{(900)^2+x^2}$$

$$25x^2=16(900)^2+16x^2$$

$$9x^2=16(900)^2$$

$$3x=\pm4(900)$$

$$x=\pm1200$$

There is no crirical number on $[0, 300].$

$$C(0)=5\sqrt{(900)^2+(0)^2}+4(300-0)$$

$$=5(900)+4(300)=5700(\$)$$

$$C(300)=5\sqrt{(900)^2+(300)^2}+4(300-300)$$

$$=5(300)\sqrt{10}=1500\sqrt{10}\approx4743.42(\$)$$

The most economical route is under the water from a power plant to a factory.