Question 329163: A museum director is trying to maximize profit. Current daily attendance is 500 people paying $30 each for a ticket. X represents the number of $5 increases in ticket price. How many $5 increases will provide the maximum income for the museum? Maximum capacity of the museum is 500 people which he meets daily.
I(x) = (500 – x) (500 + 5x)
Answer by nyc_function(2741)   (Show Source):
You can put this solution on YOUR website! I(x) = (500 - x)(500 + 5x) become y = -5x^2 + 2000x + 250,000 after multiplying the left side of the original equation.
Let y = 0 and solve for x.
Now, use the quadratic formula where a = -5, b = 2000 and c = 250,000 to solve for x. Can you take it from here?
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