F-BF Summer Intern
You have been hired for a summer internship at a marine life aquarium. Part of your job is diluting brine for the saltwater fish tanks. The brine is composed of water and sea salt, and the salt concentration is 15.8% by mass, meaning that in any amount of brine the mass of salt is 15.8% of the total mass.
- The supervisor asks you to add fresh water to one liter of the brine using a half-liter measuring cup. Let \(S(x)\) be the salt concentration of the resulting mixture when you add \(x\) half-liters of salt. Write an expression for \(S(x)\). [Assume that one liter of water has mass 1 kg.]
- Describe how the graph of \(S\) is related to the graph of \(y = 1/x\).
- Sketch the graph of \(S\).
- How much fresh water should you add to get a mixture which is 4% sea salt, approximately the salt concentration of the ocean?
Solutions
- Because the brine is 15.8% sea salt, the initial salt concentration is $$\frac{\mbox{0.158 kilograms of salt}}{\mbox{1 kilogram of brine}}.$$ Adding \(x\) half-liters of water is the same as adding \(x\) half-kilograms of mass. This increases the total mass of the mixture by the \(x\) half-kilograms, to \(0.5x+1\), while leaving the mass of salt unchanged. So an expression for \(S(x)\) is $$\frac{0.158}{0.5x+1}$$
- We write the expression for \(S(x)\) in a form that shows its relation with \(1/x\): $$\frac{0.158}{0.5x+1} = \frac{0.158}{0.5(x+2)} = \frac{0.316}{ x+2} = 0.316 \frac{1}{x+2}.$$ Thus \(S(x)\) is obtained from \(1/x\) by first replacing \(x\) with \(x+2\) and then multiplying the whole by \(0.316\). So the graph of \(S\) is the graph of $$y={1\over x}$$ horizontally translated by \(2\) units to the left, and vertically dilated by \(0.316\) units.
-
- Since a solution which is 4% sea salt has a salt concentration of \(0.04\), we must find \(x\) satisfying \(S(x) = 0.04\). So, we solve the equation \(\displaystyle{0.316\over x+2}=0.04\) for \(x\): \begin{eqnarray*} {0.316\over x+2}&=&0.04\\ 0.316&=&0.04 (x+2)\\ \frac{0.316}{0.04}&=&x+2\\ 7.9&=&x+2\\ 5.9&=&x. \end{eqnarray*} So to each liter of 15.8% brine we would have to add about 6 half-liters of fresh water (3 liters) in order to get a 4% sea salt solution.
Using a calculator or spreadsheet we construct a table of values of \(S(x)\) from the expression for \(S(x)\) found in (b):
| Number of half-liters | Salt concentration |
| 0 | 0.158 |
| 1 | 0.153 |
| 2 | 0.079 |
| 3 | 0.0632 |
| 4 | 0.0527 |
| 5 | 0.045 |
| 6 | 0.0395 |
| 7 | 0.035 |
The table shows that the salt concentration is very close to 4% when 6 half-liters (3 liters) of fresh water are added to the brine.
The solution to part (d) can be estimated from the graph found in (c) by drawing a horizontal line with vertical coordinate 0.04 and seeing what the value of \(x\) is where it intersects the graph of \(S\):
