Another revenue question?

Question

An ice cream vendor has found that the cost of supplying x cones is C(x) = 25 + 0.12x + 0.007x^2 and also that the demand for the cones increases when the price drops. Approximately 500 cones will sell per day if they are priced at $1.20, but for every price drop of 10 cents, the number sold per day increases by 20. Similarly, for every price increase of 10 cents, the number sold per day decreases by 20.
a) Determine the price of the cones that maximizes revenue.
b) Determine the price of the cones that maximizes profit.

Answer 1

Number of cones sold is

K = 740 - 2p

where p = price in cents. Also,

2p = 740 - K
p = 370 - K/2

Revenue is K * p, or

R = 370K - K^2/2

Revenue will be maximized where

0 = 370 - K
K = 370

370 cones will sell if their price is $1.85, for revenue of $684.50.

Profit is revenue minus cost. Thus,

P = 370K - K^2/2 - 2500 - 12K - .7K^2
= 358K - 1.2K^2 - 2500

Profit is maximized where

0 = 358 - 2.4K
2.4K = 358
K = 149.1666667

That many cones will sell if their price is $2.95.

Answer 2

Use a linear inequality to find both a.) and b.) its quite simple. Algebra 2