Question

5. suppose it costs $4 for a ticket to the school dance. at that price, the dance committee sells 200 tickets.

the dance committee realizes that for every $1 increase in ticket price, they sell 10 fewer tickets. find the
maximum revenue and the ticket price that maximizes the revenue.


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Answer 1

• $1440 max revenus
• $12 for a ticket

Explanation:

The quantity of tickets sold is presumed to be a linear function of the price. The revenue from sales will be the product of the number sold and their price.

__

quantity

The quantity sold is reduced by 10 when the price goes up 1, so the slope of the quantity curve is -10/1 = -10. The given point on the quantity curve is (p, q) = (4, 200). Using these values in the point-slope equation of a line, we find ...

q -200 = -10(p -4)

q = -10(p -4) +200 . . . . . add 200

revenue

Then the revenue is ...

r(p) = pq

r(p) = p(-10(p -4) +200) = p(-10p +240)

r(p) = -10p(p -24) . . . . factored form

maximum

This equation describes a parabola with zeros at p=0 and p=24. The peak value of the revenue is on the axis of symmetry, halfway between the zeros. That is, the price p=12 will maximize revenue. At that price, the revenue is ...

r(12) = -10(12)(12 -24) = 10(144) = 1440

The maximum revenue is $1440 at a ticket price of $12.00.