2020 CIME II Problems/Problem 8

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Problem 8

A committee has an oligarchy, consisting of $A\%$ of the members of the committee. Suppose that $B\%$ of the work is done by the oligarchy. If the average amount of work done by a member of the oligarchy is $16$ times the amount of work done by a nonmember of the oligarchy, find the maximum possible value of $B-A$.

Solution

Average work done sets up an equation: \[\frac{B}{A} = 16\frac{100-B}{100-A}\] \[(100-A)B = 16(100-B)A\] \[100B - AB = 1600 A - 16AB\] Let $B-A = C$ and $A+B = D$: \[50C + 50D - \frac{D^2-C^2}{4} = 800D - 800C - 4(D^2-C^2)\] \[15D^2 -3000D = 15C^2-3400C\] Complete the squares: \[15(D-100)^2 - 1500^2 = 15(C-1700/15)^2 - 1700^2/15\] \[(D-100)^2 + \frac{(17^2-15^2) \times 100^2}{15^2} = (C-1700/15)^2\] \[C = \frac{1700}{15} \pm \sqrt{(D-100)^2 + \frac{8^2 \times 100^2}{15^2}}\]

Note that $\frac{1700+800}{15}>100$ so must use minus. This means that C is maximized if $D=100$ \[C = \frac{1700}{15} - \frac{8 \times 100}{15} = 900/15=60\] $B-A$ is at a maximum $60$

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