Question 1203126: At a price of $70 for a blender, Home Outfitters will sell 12 in one month. Market research has shown that for every $5 decrease in the price of a blender, they will be able to sell 3 more each month.
a) Determine the price of a blender that will maximize revenue for the month.
b) Approximately how many blenders will be sold to reach revenue of $1100?
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52778)   (Show Source):
You can put this solution on YOUR website! .
At a price of $70 for a blender, Home Outfitters will sell 12 in one month.
Market research has shown that for every $5 decrease in the price of a blender,
they will be able to sell 3 more each month.
(a) Determine the price of a blender that will maximize revenue for the month.
(b) Approximately how many blenders will be sold to reach revenue of $1100?
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Solution for (a)
From the given information, we easily derive that 12+3n blenders
can be sold at market at the price 70-5n dollars, where n is "any" integer number,
meaning n steps of $5 decrease the initial price of $70.
Hence, the formula for revenue selling blenders at the price 70-5n dollars is
R(70-5n) = (70-5n)*(12+3n) dollars.
We want to find the value of the argument 70-5n, which provides the maximum
to the quadratic function (70-5n)*(12+3n).
For it, notice that the quadratic function (70-5n)*(12+3n), just being decomposed into
the product of linear terms, has the zeroes (= x-intecepts) at
 = 14 and  = -4.
Thus it has the maximum half-way between the x-intercepts. This half-way value is
 =  =  = 5.
So, to get an optimum price, we need to make 5 steps decreasing the initial price of $70 by $5 each time.
Thus we get the optimum price of 70-5*5 = 70-25 = 45 dollars.
At this price, 12+3n = 12+3*5 = 12+15 = 27 blenders can be solved,
providing the maximum possible revenue of 45*27 = 1215 dollars.
Compare it with the revenue of 12*70 = 840 dollars, corresponding to the initial condition.
Part (a) is solved.
Solution for (b)
To answer (b), we should find integer values n that provide the closest values of (70-5n)*(12+3n) to 1100.
To get it, I used MS Excel and prepared this Table below
n 70-5n 12+3n product
(70-5n)*(12+3n)
---------------------------------
1 65 15 975
2 60 18 1080 <<<---===
3 55 21 1155
4 50 24 1200
5 45 27 1215
6 40 30 1200
7 35 33 1155
8 30 36 1080 <<<---===
9 25 39 975
10 20 42 840
The desired values of n are n= 2 and n= 8.
The corresponding prices per blender are 70-5*2 = 60 dollars or 70-5*8 = 30 dollars (two possible values).
The corresponding amounts of blenders sold are 12+3*2 = 18 or 12+3*8 = 36 (two possible values).
Solved in full.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! revenue = price * quantity sold
when price = 70, quantity = 12 making revenue = 70 * 12 = 840.
when the price decreases by 5 dollars, the quantity sold increased by 3 dollars.
let x equal the number of times the price decreases and the quantity increases.
the equation becomes revenue = (70 - 5x) * (12 + 3x)
when x = 0, revenue = 70 * 12 = 840
when x = 1, revenue = 65 * 15 = 975
when x = 2, revenue = 60 * 18 = 1080
when x = 3, revenue = 55 * 21 = 1155
your revenue equation is y = (70 - 5x) * (12 + 3x).
simpiify to get y = 840 + 150x - 15x^2.
rearrange by descending order of degree to get -15x^2 + 150x + 840.
for the revenue to be equal to 1100, the equaton becomes:
-15x^2 + 150x + 840 = 1100
subtract 1100 from both sides of the equation to get -15x^2 + 150x + 840 -1100 = 0.
combine like terms to get -15x^2 + 150x - 260 = 0.
multiply both sides of the equation by -1 to get 15x^2 - 150x + 260 = 0.
factor the equation to get x = 7.7688746209727 or x = 2.2311253790273.
your original equation is y = -15x^2 + 150x + 840
the general form of this equation is y = ax^2 + bx + c.
y will be maximum when x = -b/(2a) = -150/-30 = 5.
when x = 5, y = -15*5^2 + 150*5 + 840 = 1215.
that's the maximum revenue.
it occurs when the price is 70 - 5*5 = 45 and the number of blenders sold is 12 + 5*3 = 27.
45 * 27 = max revenue of 1215.
this is shown on the graph at the point (5,1215).
the value of x is 5 for the maximum revenue.
this means 5 increments of 5 dollars less for the price and 5 increments of 3 units more for the quantity.
the number of increments is not an integer when you want the revenue to be exactly 1100.
the values of x that allow a revenue of at or above 1100 are 2.2311253790273 <= x <= 7.7688746209727.
if you want to make the increments integer, then you would choose x = 3 to 7.
that would make the revenue greater than 1100.
here are two displays of the graph and one display of the spreadshee i used to make the calculations.
i'll be available to answer any questions you might have.
theo
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