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When a school play charges $\$2$ for admission, an average of $100$ people attend. For each $10¢$ increase in admission price, the average number decreases by $1$. What charge would make the most money?

How do I convert this into a system of equations?

Note: For the question the system is supposed to be a parabola equation in vertex form, then find the maximum that way. For example:

A rancher is fencing off a rectangular area with a fixed perimeter of $76$m. What dimensions would yield the maximum area? Find the maximum area.

So then for this the equation would be $f(x)=3(x^2-8)+50$, solving by completing the square yields the answer of $19\times19; 361$.

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In mathematics, variables are abstract concepts that stand for concrete ideas or objects. It really does not matter what variables you use, as long as your convention is consistent and understandable.

Let us allow $P$ to stand for the profit the school makes from a play, $N$ the average number of people and $S$ the price of the tickets. We know from the problem that when the school charges \$2 per ticket (S = 2), an average of 100 (N = 100) people show up. At this price the school makes a profit of P = N $\times$ S = 2 $\times$ 100 = \$200. Now, try to figure out how changing the price of the ticket affects the profit the school makes. How does changing the price to \$2.10 affect profits, or \$2.40?

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  • $\begingroup$ I gave an example of the type of equation I need to make. $\endgroup$ – thunderbolt Jan 13 '17 at 4:27
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$2$ dollars corresponds to $100$ people. It would be wise to consider this information as the point $(2,100)$ to emphasize that they go together. Notice we chose that dollars goes with the $x$ coordinate, it doesn't matter, but we have to be consistent and logical in everything we say.

Given that dollars are corresponding to the $x$ coordinates, and the information they gave you then the slope of the graph of number of people $P$ vs the price set $x$ is $\frac{-1}{0.10}$.

I think now you know how to set up a function $P(x)$, we have a point on the line and the slope of the line.

Now that you know the amount of people that will correspond to a set price $x$, the money in hand is just,

$$xP(x)$$

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Start with the basic equation: number of people by cost. Modify the equation to add the variation of people and price. do the maximisation of the function and you get what is the maximum value

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    $\begingroup$ Yes it is, not going to lie. But homework questions aren't against site policy, and I am not even asking for the answer. $\endgroup$ – thunderbolt Jan 13 '17 at 3:56
  • $\begingroup$ See if you can follow Davids answer $\endgroup$ – unseen_rider Jan 13 '17 at 4:02
  • $\begingroup$ I am giving you hints, is more my avoidance of homework. Variation of price and profit are your only variables. Number of people(100) and cost (2) are constants and are modified by the variation of price. variation of price is equal to less 1 person. $\endgroup$ – Wilmer Ariza Jan 13 '17 at 4:04
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – thunderbolt Jan 13 '17 at 4:47
  • $\begingroup$ Your final cost will be 2+4=6 and the number of assistant will be 60 with a profit of 360 $\endgroup$ – Wilmer Ariza Jan 13 '17 at 5:44

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