SOLUTION: The demand equation for a certain product is given by p = 132 - 0.015x, where p is the unit price (in dollars) of the product and x is the number of units produced. The total reve

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The demand equation for a certain product is given by p = 132 - 0.015x, where p is the unit price (in dollars) of the product and x is the number of units produced. The total reve      Log On


   

Question 881317: The demand equation for a certain product is given by p = 132 - 0.015x, where p is the unit price (in dollars) of the product and x is the number of units produced. The total revenue obtained by producing and selling x units is given by R=xp.
Determine the prices that would yield a revenue of 9340 dollars
lowest price ___
highest price ___
I set up an equation 9430 = x(132-0.015x)
I thought I should solve for x and the substitute in to find p.
-132/(2 X -0.015) = 4400
I substituted 4400 for x in p=132-0.015x thinking I would get the a price -- $66.00 that was wrong.
I also tried to solve the above equation with the quadratic formula and couldn't come up with the answer either.
I put in 4400 and came up with 66 -- but that answer is neither the highest nor lowest price.
Found 2 solutions by Fombitz, Theo:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
9430+=+x%28132-0.015x%29
0.015x%5E2-132x%2B9430=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-132%29+%2B-+sqrt%28+%28-132%29%5E2-4%2A0.015%2A9430+%29%29%2F%282%2A0.015%29+
x+=+%28132+%2B-+sqrt%28+17424-565.8%29%29%2F%280.03%29+
x+=+%28132+%2B-+129.85%29%2F%280.03%29+
x+=+%28261.84%29%2F0.03 and x=2.16%2F0.03
x=8727.97 and x=72.02
So then,
p+=+132+-+0.015%288727.97%29
p+=+132+-+130.92
p=1.08
and
p+=+132+-+0.015%2872.02%29
p+=+132+-+1.08
p=130.92

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the solutions i came up with are:
x = 71.335 and x = 8728.6642
this was solved using the quadratic equation and also solved through graphing.

the equations that applied are:
p = 132 - .015x
that's your price equation.
r = px
that's your revenue equation.

you can substitute 132 - .015x for p in the revenue equation to get:

r = x * (132 - .015x)

simplify that to get:

r = 132x - .015x^2

you can graph that equation by substituting y for r.

the equation to graph is:

y = 132x - .015x^2

you can also graph the revenue equation of r = 9340 by graphing:

y = 9340

the intersection of these 2 equations on the graph will give you the values of x that will yield a revenue of 9340.

that graph is shown below:

picture not found

in order to solve this equation using the quadratic formula, you had to set y = 9340 in the original equation of y = 132x - .015x^2 to get:
9340 = 132x - .015x^2
subtract 9340 from both sides of this equation to get:
0 = 132x - .015x^2 - 9340 which can also be shown as:
-.015x^2 + 132x - 9340 = 0

that equation is now in standard form and you get:
a = -.015
b = 132
c = -9340

now you can apply the quadratic formula of:

               x = -b +/- sqrt(b^2 - 4ac)
                   ----------------------
                            2a


the result of those calculations will get you:

x = 71.3358 or x = 8728.66

the value of x = 4400 is the value of the axis of symmetry which leads to the maximum point on the graph when you replace x in the original equation with 4400.

that would be the equation of y = 132x - .015x^2 which becomes y = 132(4400) - .015(4400)^2.

the result of that evaluation leads to the maximum point on the graph of 290400.