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, 2017

Algebra Difficulty 5.7 AIME, harder Prove it United States

Problem:

Emilia wishes to create a basic solution with $7\%$ hydroxide $(\mathrm{OH})$ ions. She has three solutions of different bases available: $10\%$ rubidium hydroxide $(\mathrm{Rb}(\mathrm{OH}))$ , $8\%$ cesium hydroxide $(\mathrm{Cs}(\mathrm{OH}))$ , and $5\%$ francium hydroxide $(\mathrm{Fr}(\mathrm{OH}))$ . (The $\mathrm{Rb}(\mathrm{OH})$ solution has both $10\%$ $\mathrm{Rb}$ ions and $10\%$ $\mathrm{OH}$ ions, and similar for the other solutions.) Since francium is highly radioactive, its concentration in the final solution should not exceed $2\%$ . What is the highest possible concentration of rubidium in her solution?

Solution

Solution:

Suppose that Emilia uses $R$ liters of $\mathrm{Rb}(\mathrm{OH})$ , $C$ liters of $\mathrm{Cs}(\mathrm{OH})$ , and $F$ liters of $\mathrm{Fr}(\mathrm{OH})$ , then we have
$\frac{10\% \cdot R + 8\% \cdot C + 5\% \cdot F}{R + C + F} = 7\% \text{ and } \frac{5\% \cdot F}{R + C + F} \leq 2\%$
The equations simplify to $3R + C = 2F$ and $3F \leq 2R + 2C$ , which gives
$\frac{9R + 3C}{2} \leq 2R + 2C \Rightarrow 5R \leq C$
Therefore the concentration of rubidium is maximized when $5R = C$ , so $F = 4R$ , and the concentration of rubidium is
$\frac{10\% \cdot R}{R + C + F} = 1\%$

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