word problem for math; please help??!?

Question

i dont get this, please help me:

a store sells about 50 of a new model of dvd player per month at a price of $140 each. for each $10 decrease in price, about 5 more dvd players per month are sold. how much should the store charge in order to maximize monthly revenue? what is the maximum monthly revenue?

Answer 1

Let
p = unit price
n = units sold at price p
R = revenue = np
(n - 50) / (p - $140) = - 5/$10
n - 50 = - 1(p - $140) / $2 = 70 - p/2
n = - p/2 + 120
R = - p^2/2 + 120p
R = - 1/2(p^2 - 240p + 120^2) + 7,200
R = - 1/2(p - $120) + $7200
Optimum unit price = $120
Maximum monthly revenue = $7,200
from sale of 60 units/month.

Answer 2

If P = the price in dollars, then the number of DVD players sold is N = 50 + 5 * (140 - P)/10 = 120 - P / 2.

Revenue is N * P = 120 P - P^2 / 2.

Take the derivative of the revenue with respect to P to get 120 - P. Set this expression to zero and solve for P to get P = 120. So, a price reduction of $20 will maximize revenue.

The number of units sold would be 60, so revenue would be $120 * 60 = $7,200.

Answer 3

Okay, look at it this way:

The store has 50 DVD players at $140 each:
After the price drop (@$10) the revenue is ($140-$10)*5=$650. Now the store has 45 DVd players left.
After another price drop (@$10) the revenue is ($130-$10)*5=$600. Now the store has 40 DVD player left.
Continue 8 more time until the number of DVD player reach 0. Have a question please email me at [e-mail].

Answer 4

this is a maxima problem. the hard part is setting up the equation

equation is revenue = price * number sold

revenue= (140+x)(50 - .5x)

this means where for example x is ten so you pay 150. number sold will go down to 45. if x is 20 number sold goes to 40 etc. and vice versa if x is negative

simplify equation then differentiate and set it equal to zero
solve