calculusoptimization
A power house, P, is on one bank of a straight river $200$ m wide,
and a factory, F, is on the opposite bank $400m$ downstream from P.
The cable has to be taken across the river,
under water at a cost of Rs $6/m$.
On land the cost is Rs $ 3/m$.
What path should be chosen so that the cost is minimized?
I first looked at the extreme cases and then tried to find the solution. But then too there was a problem
The picture for this situation looks like this:
The cost function is that given in the comment attached to angryavian's answer. A discussion for a similar optimization problem (of a very commonly-assigned sort) can be found here. Differentiating the cost function to find its critical point gives us
$$ \frac{dC}{dx} \ = \ 25000 \ \left( \ 1 \ + \ 2 \ \cdot \frac{1}{2} \ [ \ 2^2 \ + \ (3-x)^2 \ ]^{-1/2} \ \cdot \ 2 \ (3-x) \ \cdot (-1) \right) \ = \ 0 $$
$$ \Rightarrow \ \ \frac{2 \ (3-x)}{\sqrt{2^2 \ + \ (3-x)^2}} \ = \ 1 \ \ \Rightarrow \ \ 3 \ - \ x \ = \ \sqrt{\frac{4}{3}} \ \ \Rightarrow \ \ x \ \approx \ 1.85 \ \text{km.} \ \ . $$
This makes the minimal total cost
$$ C(1.85) \ \approx \ 25000 \ \cdot \ 1.85 \ + \ 50000 \ \sqrt{4 \ + \ (3 -1.85)^2} \ \approx \ 161,700 \ \text{dollars} \ \ . $$
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 $$\min_{l, r}l+2r$$
Best Answer
Draw a picture. Let $A$ be the point directly across the river from $P$.
Suppose we go underwater directly to a point $x$ metres downstream from $P$. Then by the Pythagorean Theorem, we will have a $\sqrt{x^2+200^2}$ underwater stretch. Then we need to travel $400-x$ overland to $F$. The cost $C(x)$ is given by $$C(x)=6\sqrt{x^2+200^2}+3(400-x).$$ To minimize, find the derivative $C'(x)$ and set it equal to $0$. We have $$C'(x)=\frac{6x}{\sqrt{x^2+200^2}}-3.$$ Solving $C'(x)=0$ is easy. We get $2x=\sqrt{x^2+200^2}$. Square both sides.
When you have done this, and found a candidate $x$, you need to check whether this really gives the minimum cost.
Remark: There are various other ways to solve the problem. We chose the most basic beginning calculus approach.