SOLUTION: A concession stand sells drinks at an event. When they sell the drinks for $1.50, they sell approximately 3,000. When they increased the price to $1.75, they only sold about 2,500

Question 1114400: A concession stand sells drinks at an event. When they sell the drinks for $1.50, they sell approximately 3,000. When they increased the price to $1.75, they only sold about 2,500 (assume linear). If it costs $0.40 to purchase each drink and $1,000 per event to rent the space...
1) How much should they sell the drinks for in order to maximize their profits?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

When they sell the drinks for $1.50, their profit is $1.50-$0.40 = $1.10. At that price, they sell 3000 drinks.

When they raise the price by $0.25, the number of drinks they sell drops by 500 to 2500. We are to assume that relationship is linear -- i.e., raising the price another 25 cents will result in an addition reduction of 500 in the number of drinks sold.

So if x is the number of times the price is increased by 25 cents, then the profit is
(profit per drink) times (number of drinks) minus (fixed (rental) cost):

This is a quadratic equation; the maximum value is when

The maximum profit will be achieved if they raise the price by 25 cents 0.8 times -- i.e., if they raise the price by 20 cents, to $1.70.

Answer: The price at which they should sell the drinks to maximize profit is $1.70.
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