What is the Minimal cost? A power house, $P$, is on one bank of a straight river $W$ meters (m) wide, and a factory, $F$, is on the opposite bank $L$ meters downstream from $P$. The cable has to be taken across the river, under water at a cost of $\$CW/\mathrm{m}$. On land the cost is $\$CL/\mathrm{m}$. What is the minimum cost?
Any help will be appreciated...
 A: There are two ways of routing the cable - 


*

*Completely through water (if CW is not too high than CL) and 

*Maximum by land and then only the width of the river by water (if CW is too high than CL). 


The costs are respectively C1= {(L^2+W^2)^.5}*CW {using pythagorus theorem} and C2= (CW*W + CL*L)
Hence, minimum cost = Min(C1,C2).
A: Set this up as a non linear programming problem. 
1.Variables:
$x_L=$the meters of cable to be used on land, 
$x_W=$the meters of cable to be used under water
2.Objective Function: 
You want to minimize the total cost that is $$z=\min_{x_L,x_W}CL\cdot x_L + CW \cdot x_W$$
3.Restrictions
$$L^2+W^2=x_{L}^2+x_{W}^2-2x_Lx_w\cos(\gamma)$$
where $\gamma$ is the angle between the sides $x_L$ and $x_W$. The explanation of this condition is the following. You have a triangle with edges $P$, $F$ and another edge, say $D$ at the point where your cable meets the coast where $F$ is. Now, consider the right angle triangle with sides $W$, $L$ and $\sqrt{L^2+W^2}$. (Do you see it?). Begin to move edge $D$ towards $F$. The new triangle(s) has(ve) not sides  $W$, $L$ and $\sqrt{L^2+W^2}$ anymore, but $x_W$, $x_L$ and $\sqrt{L^2+W^2}$ instead. The side that remains unaffected is the straight line between $P$ and $F$. 
Note now that you have also an angle that is constant and can be determined. Consider the following description. As already mentioned, you have a staight line (the hypotenuse) from P to F. The length  of this diagonal is equal to $L^2+W^2$ due to Pythagoras law (draw a shape). Then you can move along the coast and reduce $x_L$ but increase $x_W$. Note though, that the angle between the coast line (where F is) and the diagonal remains unaffected. Call this angle $\alpha$. In the right angle triangle with sides $W$, $L$ and $\sqrt{L^2+W^2}$ we can easily calculate $\alpha$, since $\sin(\alpha)=\frac{W}{\sqrt{L^2+W^2}}$. Then the law of sines implies that $$\frac{\sqrt{L^2+W^2}}{\sin(\gamma)}=\frac{x_W}{\sin(\alpha)}$$ Solving for $\sin(\gamma)$ and by substituting we find that $$\sin(\gamma)=\frac{W}{x_W}.$$ Therefore $$\cos(\gamma)=\sqrt{1-\sin^2(\gamma)}=\sqrt{1-\frac{W^2}{x_W^2}}.$$
So you have the nonlinear restriction
(1) $L^2+W^2=x_{L}^2+x_{W}^2-2x_Lx_w\sqrt{1-\frac{W^2}{x_W^2}}$
and the restrictions:
(2) $x_L \ge 0$ (nonnegativity condition)
(3) $x_W \ge W$ (you have to cross the river)
To sum up: 
$$z=\min_{x_L,x_W}CL\cdot x_L + CW \cdot x_W$$
under the restrictions
(1) $L^2+W^2=x_{L}^2+x_{W}^2-2x_Lx_w\sqrt{1-\frac{W^2}{x_W^2}}$
(2) $x_L \ge 0, \, x_W \ge W$
Now use Excel NLP solver to solve it, or KKT conditions if you have the time.
