Answer to Question #148337 in Calculus for Angelo

Question

Points A and B are opposite each other on shores of a straight river 3 km wide. Point C is on the same shore as B but 2 km down the river from B. A telephone company wishes to lay a cable from A to C. If the cost per kilometer of the cable is 25% more under the water than it is on land, what line of cable would be least expensive for the company? (Let

Answer 1

Let`s solve this problem:

We have given lengths:

From A to B = 3 km, from B to C = 2 km.

Let`s assume that the value of 1 km cable in water is x and its value in land is y. And we have also an important thing that we need to assume first: location of C is 2 km down from B, it means that it can be under water or in land because location of B did not given, just mentioned that located on shore. But in this case, we asseume that C located in land.

We need to look two or more cases

1) First, we will find the value of the cable which used in the direction of A to B directly and C:

From A to B $3\times x$ is needed and From B to C $2\times y$ and its total value is $3\times x$+$2\times y$.

2) Secondly, we consider the value of cable from A to C directly:

we can calculate the length from Pythagorean theorem:

From A to C = $\sqrt{\smash[b]{3^2+2^2}}=\sqrt{\smash[b]{13}}$ and it value is $\sqrt{\smash[b]{13}}\times x$.

At last, we have given connection between x and y:

$x=y+0.25\times y=1.25\times y$ and in order compare them, we put x into equations:

this is from A to C through B: $3\times x$+$2\times y$=$3\times 1.25\times y+2\times y=5.75\times y$,

this is fromA to C directly: $\sqrt{\smash[b]{13}}\times x=\sqrt{\smash[b]{13}}\times 1.25\times y=4.50\times y$.

3) In the last case, we put D point between B and C:

From A to D: $\sqrt{\smash[b]{3^2+1^2}}=\sqrt{\smash[b]{10}}$ m cable required and its value $\sqrt{\smash[b]{10}}\times x= \sqrt{\smash[b]{10}}\times 1.25\times y=3.95\times y$

So, I think that if a telephone company lays cable from A to D directly, then it will be cheaper