Minimizing the cost of a pipeline over land and water. 
An oil refinery is located on the north bank of a straight river that is $2$ km wide. A pipeline is to be constructed from the refinery to storage tanks on the south bank of the river $6$ km east of the refinery. The cost of laying pipe is $\$400,000$ per km over land and $\$800,000$ per km under the river to the tanks. How should the oil company lay the pipe in order to minimize the cost? What is the minimum cost? 

So I got x=2.617 once I found my derivative and solved. So to receive the cost would I multiple x by the two amounts?
 A: Straight line (diagonally through the river) from the refinery to the storage tanks is due to Pythagoras theorem $$2^2+6^2=4^2$$ equal to $4$km. Any other connection forms a triangle with sides $l$ (part of pipeline on land in km) and $r$ (part of pipeline below the river in km), so that $l^2+r^2-2lr\cos{\theta}=4^2$. So, you want to minimize the cost $$\min_{l, r}400,000l+800,000r$$ subject to the constraint $$l^2+r^2-2lr\cos{\theta}=4^2$$ where $θ$ is the angle between the sides $l$ and $r$, with $0^o\le θ\le 90^o$. The objective funcion can be equivalently restated as $$\min_{l, r}l+2r$$
A: Let $x$ be the length of pipe that is laid to the east of the refinery over land, where $x \in [0, 6]$. Then by forming a right triangle, it follows by Pythagoras that the length of pipe that is laid under the river is given by the expression:
$$
\sqrt{2^2 + (6 - x)^2}
$$
Thus, the desired objective function to be minimized is:
$$
f(x) = 400000x + 800000\sqrt{4 + (6 - x)^2}
$$
A: Remembering that Snell's law involves the angle between a beam of light and the normal to the point of incidence, it may seem advantageous to express everything in terms of an unknown angle rather than an unknown length.
To do this, cross the river first at an angle $x$ with the direct route across and then go along the bank the remaining distance (it doesn't matter which bank you go along, the cost is the same). Take $\$400,000$ as one unit of money to lose unnecessary zeros etc in the calculation.
The distance across the water is $\cfrac 2{\cos x}$km at a cost of $\cfrac 4{\cos x}$ units. The distance on land is $\left(6 - 2 \tan x\right)$km at a cost of $\left(6 - 2 \tan x\right)$ units.
The total cost is $\left(6 - 2\tan x+\cfrac 4{\cos x}\right)$ units

Using $\left(\cfrac uv\right)'=\cfrac {u'v-v'u}{v^2}$ with the derivative of $\sin x$ being $\cos x$ and the derivative of $\cos x$ being $-\sin x$ we find:
The derivative of $\tan x$ is $\left(\cfrac {\sin x}{\cos x}\right)'=\cfrac {\cos^2x+\sin^2x}{\cos^2 x}=\cfrac 1{\cos^2 x}$ 
And $\left(\cfrac 1{\cos x}\right)' = \cfrac {\sin x}{\cos^2 x}$
