Calculus 1 Courses's Optimization of Pipeline Route Problem, Produces Strangely Complicated Value.

Question

I am a Calculus 1 student and I have an optimization word-problem that is giving me a lot of trouble.

It has two variables. I have found the value for $y$, but when I plugged it into the equation and tried to solve for the $x$ I couldn't find it's value. I used Symbolab to solve it, but it came up with a decimal number that's extremely complicated when written as a fraction. My professor has given us very complicated problems before, but the complexity of this number is such that I feel like it's very likely I did something wrong.

I have checked other parts of my work with Symbolab and I am still not sure where I went wrong, but I would really appreciate it if you would take a look and determine if there are any parts that don't look right to you.

---

An oil refinery is located on the north bank of a straight river which is $2$km wide. A pipeline is to be constructed from the refinery to storage tanks located 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 to a point $P$ on the north bank and $\$800,000$ per km under the river to the tanks. To minimize the cost of the pipeline, where should $P$ be located?

$P=$ The area where the pipeline enters the river.

$x=$ The horizontal distance between the oil refinery and the storage tanks.

$y=$ The euclidean distance between $P$ and the storage tanks.

The Pythagorean theorem states $2^2+(6-x)^2=y^2$.

The cost of the pipeline is $C = 400,000x+800,000y$.

---

finding $y$:

$$4+(6-x)^2 = y^2 \to y= \pm \sqrt{4+(6-x)^2}$$

Differentiating $C = 400,000x+800,000\sqrt{4+(6-x)^2}$:

$$\frac{d}{dx}\sqrt{4+(6-x)^2}=\frac{1}{2\sqrt{4+(6-x)^2}}\cdot-2(6-x)=\frac{-(6-x)}{\sqrt{4+(6-x)^2}}$$

$$\frac{d}{dx}800,000\sqrt{4+(6-x)^2}=[\frac{-(6-x)}{\sqrt{4+(6-x)^2}}\cdot800,000] = \frac{-800,000(6-x)}{\sqrt{4+(6-x)^2}}$$

$$\frac{d}{dx}400,000x+800,000\sqrt{4+(6-x)^2}=400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}}$$

Setting $400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$ and solving for $x$.

$$x=4.84530..$$

I'm not completely sure how to write the fraction out with math notation here because a single square root seems to cover part of the numerator and all of the denominator, but you can see it if you plug $400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$ into Symbolab's "solve for" calculator.

Answer 1

As a commenter has made clear, when solving $400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$ for $x$, $x=6-\frac{2}{\sqrt{3}}$.

Steps Involved:

$$400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$$

Subtract $400,000$ from both sides.

$$\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 400,000$$

Divide both sides by $400,000$.

$$\frac{2(6-x)}{\sqrt{4+(6-x)^2}}=1$$

Multiply both sides by $\sqrt{4+(6-x)^2}$.

$$2(6-x)=\sqrt{4+(6-x)^2}$$

Take both sides to the power of $2$

$$4(6-x)^2=4+(6-x)^2$$

Subtract $(6-x)^2$ from both sides.

$$3(6-x)^2=4$$

Divide both sides by $3$.

$$(6-x)^2=\frac{4}{3}$$

Square both sides.

$$6-x=\frac{2}{\sqrt{3}}$$

Subtract $6$ from both sides.

$$x=6-\frac{2}{\sqrt{3}}$$