I been trying to solve this question and have tried to solve it for many days, but do not know how, any help would be much oblidged.
A cable company owns the roads marked with the dotted lines in the figure, and it costs the cable company 25 dollars per foot to run the line on poles along the county roads. The area bounded by the house and roads is a privately owned field, ant the cable company must pay for an easement to run lines underground in the field. It is also more costly for the company to run the line underground than to run them on the poles. The total cost to run lines underground across the field is $52 per foot. Whilst to run the cable via pole 46% less than to run it underground.
Image: https://i.stack.imgur.com/2A5ib.jpg
The cable company has the choice of running the lines line along the road or cutting across the field.
I. Advise the company on the costings to the company to run lines both along the road to H and directly across the field.
II. What would the critical distance that the piped must be laid which will contribute to a minimum costings to the company.
For part I, I had used pythagor's theorem to find the distance of PH and multiply it by the respective costs of the material.
However for II, I am a bit stumped. The critical distance is defined as the distance from p to the second point on the line, let us call it G. So critical distance is equal to PG, however, I do not know how to find it since the value of that hypotenuse is not given, and the height is ewual to (100-x).
Plugging in this value in the theorem gives an unimaginably large expression without any real roots. Any help would be much oblidged.
Let u be the cost (per unit) for underground.
Let p be the cost (per unit) for using pole. C the total cost.
Then, $C = u \sqrt {125^2 + x^2} + p(100 – x)$
$\frac {dC}{dx} = \frac {ux}{\sqrt {125^2 + x^2}} – p$
$\frac {dC}{dx} = 0$ implies $\frac {ux}{\sqrt {125^2 + x^2}} = p$
From the above, $x$ can be found and is $\frac {125p}{\sqrt {u^2 - p^2}}$
Note: I have skipped the second derivative test to make sure that the cost is a minimum.