Game Show Score Maximization
Sarah, James, and Matthew are on a team in a game show. In the game, Sarah always earns 5 points, Matthew always earns -2 points, and James always earns 3 points. To calculate their team's score, they multiply the first person's score by the second person's score, then subtract the third person's score. If they can choose the order in which they play the game, what is the maximum possible score that their team can earn?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
If we let the first, second, and third persons' score equal $ a $, $ b $, and $ c $ respectively, then their team's score is $ ((a \cdot b)-c) $. We may test all six possible orderings.
$ a = 5 $, $ b = -2 $, $ c = 3 \rightarrow ((5 \cdot (-2)) - 3) = -13 $.
$ a = -2 $, $ b = 5 $, $ c = 3 \rightarrow (((-2) \cdot 5) - 3) = -13 $.
$ a = 5 $, $ b = 3 $, $ c = -2 \rightarrow ((5 \cdot 3) - (-2)) = 17 $.
$ a = 3 $, $ b = 5 $, $ c = -2 \rightarrow ((3 \cdot 5) - (-2)) = 17 $.
$ a = 3 $, $ b = -2 $, $ c = 5 \rightarrow ((3 \cdot (-2)) - 5) = -11 $.
$ a = -2 $, $ b = 3 $, $ c = 5 \rightarrow (((-2) \cdot 3) - 5) = -11 $.
Thus, our total answer is $ \boxed{17} $
Additionally, we may note that the only way to obtain a positive answer is to multiply the two positive scores together and subtract the negative score. Because multiplication is commutative, the order of $ a $ and $ b $ does not matter, and we need only compute the single case $ (3 \cdot 5) - (-2) = \boxed{17} $
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