Question

A power line is to be constructed from a power station at point A to an island at point​ C, which is 2 mi directly out in the water from a point B on the shore. Point B is 8 mi downshore from the power station at A. It costs ​$4200 per mile to lay the power line under water and ​$3000 per mile to lay the line under ground. At what point S downshore from A should the line come to the shore in order to minimize​ cost? Note that S could very well be B or A. ​(Hint​: The length of CS is StartRoot 4 plus x squared EndRoot​.)

Answer 1

Answer:

The distance downshore from point A to point S that minimizes cost is 5.5 miles.

Step-by-step explanation:

We can define the value X as the length that the power line is lie underground across the shore.

Then the segment Y becomes:

$Y^2=2^2+(8-X)^2\\\\Y^2=4+64-16X+X^2=70-16X+X^2\\\\Y=\sqrt{70-16X+X^2}$

Then, we can write the cost of the line (Z) as:

$Z=3000X+4200Y=3000X+4200\sqrt{70-16X+X^2}$

To optimize the cost function, we derive and equal to zero:

$\dfrac{dZ}{dX}=3000(1)+4200[\dfrac{1}{2}(70-16X+X^2)^{(1/2-1)}\cdot(-16+2X)]\\\\\\ \dfrac{dZ}{dX}=3000+4200\dfrac{2X-16}{2\sqrt{70-16X+X^2}}=0\\\\\\4200\dfrac{2X-16}{2\sqrt{70-16X+X^2}}=-3000\\\\\\\dfrac{2X-16}{2\sqrt{70-16X+X^2}}=-3000/4200=-5/7\\\\\\\dfrac{X-8}{\sqrt{70-16X+X^2}}=-5/7\\\\\\X-8=-\dfrac{5}{7}\sqrt{70-16X+X^2}\\\\(X-8)^2=\dfrac{25}{49}(70-16X+X^2)\\\\X^2-16X+64=\dfrac{1750-400X+25X^2}{49}\\\\\\49X^2-784X+3136=1750-400X+25X^2\\\\24X^2-384X+1386=0\\\\4X^2-64X+231=0$

$X_{1,2}=\dfrac{-(-64)\pm\sqrt{(-64)^2-4(4)(231)}}{2(4)}\\\\\\X_{1,2}=\dfrac{64\pm\sqrt{4096-3696}}{8}\\\\\\X_{1,2}=\dfrac{64\pm\sqrt{400}}{8}=\dfrac{64\pm20}{8}\\\\\\X_1=\dfrac{64-20}{8}=\dfrac{44}{8}=5.5\\\\\\X_2=\dfrac{64+20}{8}=\dfrac{84}{8}=10.5\rightarrow\text{invalid: X}\leq 8$

The distance downshore from point A to point S that minimizes cost is 5.5 miles.