Question 1114400
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When they sell the drinks for $1.50, their profit is $1.50-$0.40 = $1.10.  At that price, they sell 3000 drinks.<br>
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.<br>
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):
{{{(1.10+.25x)(3000-500x)-1000 = 3300+750x-550x-125x^2-1000 = -125x^2+200x+2300}}}<br>
This is a quadratic equation; the maximum value is when {{{x = (-200)/(2(-125)) = (-200)/-250 = 0.8}}}<br>
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.<br>
Answer: The price at which they should sell the drinks to maximize profit is $1.70.