Thread: 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).

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.

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

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

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??

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)