Question 446766
An amusement park has been charging $12 per person for admission and averaging 1000 patrons per day.
 the directors of the park are considering a $2 increase in the price for next season .
 They have calculated that for each $2 in increase they will probably lose 100 patrons per day.
 What should be the ideal admission fee for the greatest income on an average day
 . Write an equation to model the situation and draw a graph of the situation to justify your answer.
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I think you are right, but I would do it this way.
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Let x = no. of $2 increases and no. of 100 patron decreases
:
Income = price * no. patrons
f(x) = (12+2x)*(1000-100x)
Foil
f(x) = 12000 - 1200x + 2000x - 200x^2
A quadratic equation
f(x) = -200x^2 + 800x + 12000, is the equation
:
The greatest income occurs when x = the axis of symmetry, x = -b/(2a)
x = {{{(-800)/(2*-200)}}}
x = +2
Therefore max income when
12+2(2) = $16 is the price (which you got)
then
1000 - 2(100) = 800 patrons at that price
Max income
16 * 800 = $12,800 
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To graph this
{{{ graph( 300, 200, -4, 8, -2000, 18000, -200x^2+800x+12000) }}}
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You can confirm this, substitute 2 in the equation to find f(x) = 12,800