math! math! crisis! please help!?

Question

a shoe store sells about 200 pairs of a new basketball shoe each month when it charges $60 per pair.For each $1 increase in price, about 2 fewer pairs per month are sold.

How much per pair should the store charge to maximize monthly revenue?
You: what is the maxium revenue?

the answers are $80; $12800.
The point is I don't know how to get the answers!
please give me explanation not just the answer! thank you!

Answer 1

So revenue = price times number sold
= (200 - 2p)(60 + p) where p = additional price charged

This FOILed out is 12000 + 80 p - 2p^2

Its maximum occurs at the axis of symmetry, p = -80 over 2(-2) which is 20.

So increase $20- plug in to get max. revenue

Answer 2

You need to obtain a relationship between the selling price and the number of pairs of shoes sold. You're told that when the price is $60, 200 pairs/month are sold. and that a rise of $1 in the selling price reduces the sales by 2 pairs/month. So, for example, if the price is $61, sales would be 198 pairs/month; if the price is $62, sales would be 196 pairs/month; and so on. This is a linear relationship. If p is the selling price and s is the number of pairs of shoes sold per month, you could start with

s = a*p + b

where a and b are constants to be found. We know that when p=60, s=200, and that when p=61, s=198. So

200 = a*60 + b
198 = a*61 + b

It is easy to solve this; you get a = -2, b = 320. So the sales per month, as a function of price, is given by

s = -2p + 320

Now,

Revenue = (selling price)*(sales)

so

R = p*(-2p + 320) = -2p² + 320p

Maximize R.

Answer 3

This is an optimization problem.

Let
p = price of pair of shoes
x = number of dollars above $60 the price of shoes is
n = number of pairs of shoes sold
R = revenue

R = np = (200 - 2x)(60 + x)
R = 12,000 + 80x - 2x²

If this is a NOT a calculus class find the vertex of the parabola.
If this IS a calculus class take the derivative and set it equal to zero to find the critical points.

dR/dx = 80 - 4x = 0
4x = 80
x = 20

p = 60 + x = 60 + 20 = 80
n = 200 - 2x = 200 - 2*20 = 200 - 40 = 160

R = np = 80*160 = $12,800

So maximum revenue of $12,800 is obtained by selling 160 pairs of shoes at $80 per pair.

Answer 4

An explanation is to calculate:
Is the dollar gained for each pair worth the loss of two sales:
A way of calculating sales lost for the dollar amount
P = pairs lost I = Price Increase D = Pairs lost per dollar
P = ID
A way of calculating revenue for each price:
R = Revenue P = Price S = Pairs sold
R = PS
At first you gain money has you increase price:
$60 200 pairs:
$12,000 = ($60)(200)
$70 180 pairs:
$12,600 = ($70)(180)
$79 162 pairs:
$12,798 = ($79)(162)
Maximum revenue:
$80 160 pairs
$12,800 = ($80)(160)
Then you start losing it:
$81 158 pairs
$12,978 = ($81)(158)

Answer 5

if $I is the increase in price then the cashpermonth = (60+I)x(200-2I) dollars. The maximum of this quadratic occurs at I=20 thus the answer is $80. No guess work

Answer 6

alright let me work this baby out...youre gonna want a function, so make a table putting values in for what x (the price) is, and for y (how many pairs are sold) ok so when x=60, y=200, when x=61 y=198 and when x=62 y=196, idk if you know how to change that table into a function but what youre eventually gonna get is y= (-2)x+320 use that to find a value for x that you can multiply by the resulting y to equal the highest value...phew...or you can enter the function into a graphing calculator and find the max..

Answer 7

best bet is to use TRIAL AND ERROR. Estimate an answer and plug it in, then go one higher or one lower until you get the MAX result. Start maybe with 75, then try 76, 77--etc when you do 80 you get your answer and when you do 81, the result is lower, so you know 80 is the right answer.