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Difference between revisions of "2020 CIME II Problems/Problem 8"

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<cmath> C = \frac{1700}{15} - \frac{8 \times 100}{15} = 900/15=60</cmath>
 
<cmath> C = \frac{1700}{15} - \frac{8 \times 100}{15} = 900/15=60</cmath>
 
<math>B-A</math> is at a maximum <math>60</math>
 
<math>B-A</math> is at a maximum <math>60</math>
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==Solution 2==
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As in the first solution, we get <math>100B - AB = 1600A - 16AB</math>. We rearrange and obtain <math>15AB + 100B - 1600A = 0</math>. We divide by <math>15</math> to obtain <math>AB + \frac{100}{15}B - \frac{1600}{15}A = 0</math>. We then subtract <math>\frac{160000}{225}</math> from both sides, and factor to obtain <math>(A + \frac{100}{15})(B - \frac{1600}{15}) = -\frac{1600000}{225} = -\left(\frac{400}{15}\right)^2</math>. If we graph this with <math>A</math> being on the <math>x</math>-axis and <math>B</math> being on the <math>y</math>-axis, this equation is the hyperbola <math>xy = 1</math>, except scaled up by <math>\frac{400}{15}</math> and translated <math>\frac{100}{15}</math> to the left and <math>\frac{1600}{15}</math> up. This graph intersects <math>(0, 0)</math> and <math>(100, 100)</math>, and the maximum difference clearly occurs at the point when the slope of the function is <math>1</math>. This is at <math>(-\frac{100}{15} + \frac{400}{15}, \frac{1600}{15} - \frac{40}{15}) = (20, 80)</math>. Our answer is <math>\boxed{60}</math>.
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~mathboy100

Revision as of 16:15, 6 February 2023

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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