# Math Help - Maximizing revenue

1. ## Maximizing revenue

A smoothie stand usually sells 150 smoothies per day at $4 each. a business student's research tells her that for every$0.10 decrease in the price, the stand will sell 5 more smoothies per day. Using calculus determine the price at which the smoothies should be sold to maximize revenue. Be very clear when defining your variable(s).

2. Originally Posted by spencersin
A smoothie stand usually sells 150 smoothies per day at $4 each. a business student's research tells her that for every$0.10 decrease in the price, the stand will sell 5 more smoothies per day. Using calculus determine the price at which the smoothies should be sold to maximize revenue. Be very clear when defining your variable(s).
Let x represents the number of times the price decrease.

Price of each smoothie = $(4 - 0.10x) Number of smoothies = (150 + 5x) Revenue = (150 + 5x)(4 - 0.10x) Now maximize Revenue using calculus. 3. Originally Posted by spencersin A smoothie stand usually sells 150 smoothies per day at$4 each. a business student's research tells her that for every $0.10 decrease in the price, the stand will sell 5 more smoothies per day. Using calculus determine the price at which the smoothies should be sold to maximize revenue. Be very clear when defining your variable(s). Revenue = (# of smoothies sold)*(price at which they are sold) Now, the revenue $r(x) = (150 + 5x)(4 - 0.1x)$ since for each$0.1 we take off the price of $4, we increase the number sold by 5 times that number. your goal is to maximize $r(x)$ 4. I found that R'(x) = 5-x So, x=5 <---critical number how can I get the price at which the smoothies should be sold to maximize revenue?? 5. Originally Posted by spencersin I found that R'(x) = 5-x So, x=5 <---critical number how can I get the price at which the smoothies should be sold to maximize revenue?? ok, is this a maximum point for R(x)? remember what price you want to sell at,$(4 - 0.1x)