Difference between revisions of "2020 CIME II Problems/Problem 8"
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<cmath> C = \frac{1700}{15} \pm \sqrt{(D-100)^2 + \frac{8^2 \times 100^2}{15^2}}</cmath> | <cmath> C = \frac{1700}{15} \pm \sqrt{(D-100)^2 + \frac{8^2 \times 100^2}{15^2}}</cmath> | ||
− | Note that <math>\frac{1700+800}{15}> | + | Note that <math>\frac{1700+800}{15}>100</math> so must use minus. This means that C is maximized if <math>D=100</math> |
− | <cmath> C = \frac{1700}{15} - \frac{8 \times 100}{15 | + | <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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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>. | ||
+ | |||
+ | ~mathboy100 | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | Assume WLOG, there are <math>100</math> members and that there are <math>100</math> work that is done. So, there are <math>A</math> members of the oligarchy and they've done <math>B</math> work in total. Thus, on average, each member of the oligarchy does <math>\dfrac{B}{A}</math> work. Then, the average work a nonmember of the oligarchy will be <math>\dfrac{B}{16A}</math>. Thus, | ||
+ | <cmath>B+(100-A)\cdot \dfrac{B}{16A} = 100</cmath> | ||
+ | <cmath>\implies 16AB+(100-A)B = 1600A</cmath> | ||
+ | <cmath>\implies 3AB+20B-320A = 0</cmath> | ||
+ | <cmath>\implies (3B-20)(3A+20) = -6400</cmath> | ||
+ | When <math>3B-20 = -80</math> and <math>3A+20 = 80</math>, we have <math>A=20, B=80</math>, giving <math>B-A = 80-20 = \boxed{60}</math> | ||
+ | |||
+ | ~sml1809 |
Latest revision as of 17:36, 28 December 2023
Contents
Problem 8
A committee has an oligarchy, consisting of of the members of the committee. Suppose that
of the work is done by the oligarchy. If the average amount of work done by a member of the oligarchy is
times the amount of work done by a nonmember of the oligarchy, find the maximum possible value of
.
Solution
Average work done sets up an equation:
Let
and
:
Complete the squares:
Note that so must use minus. This means that C is maximized if
is at a maximum
Solution 2
As in the first solution, we get . We rearrange and obtain
. We divide by
to obtain
. We then subtract
from both sides, and factor to obtain
. If we graph this with
being on the
-axis and
being on the
-axis, this equation is the hyperbola
, except scaled up by
and translated
to the left and
up. This graph intersects
and
, and the maximum difference clearly occurs at the point when the slope of the function is
. This is at
. Our answer is
.
~mathboy100
Solution 3
Assume WLOG, there are members and that there are
work that is done. So, there are
members of the oligarchy and they've done
work in total. Thus, on average, each member of the oligarchy does
work. Then, the average work a nonmember of the oligarchy will be
. Thus,
When
and
, we have
, giving
~sml1809