2020 CIME II Problems/Problem 8

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 $60$

Solution 2

As in the first solution, we get $100B - AB = 1600A - 16AB$ . We rearrange and obtain $15AB + 100B - 1600A = 0$ . We divide by $15$ to obtain $AB + \frac{100}{15}B - \frac{1600}{15}A = 0$ . We then subtract $\frac{160000}{225}$ from both sides, and factor to obtain $(A + \frac{100}{15})(B - \frac{1600}{15}) = -\frac{1600000}{225} = -\left(\frac{400}{15}\right)^2$ . If we graph this with $A$ being on the $x$ -axis and $B$ being on the $y$ -axis, this equation is the hyperbola $xy = 1$ , except scaled up by $\frac{400}{15}$ and translated $\frac{100}{15}$ to the left and $\frac{1600}{15}$ up. This graph intersects $(0, 0)$ and $(100, 100)$ , and the maximum difference clearly occurs at the point when the slope of the function is $1$ . This is at $(-\frac{100}{15} + \frac{400}{15}, \frac{1600}{15} - \frac{40}{15}) = (20, 80)$ . Our answer is $\boxed{60}$ .

~mathboy100

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2020 CIME II Problems/Problem 8

Problem 8

Solution

Solution 2