Question 1109235: A candy company was looking at its profits and wanted to know the maximum number of candies that needed to be sold to make a maximum profit. Write a possible equation where it is easy to find the maximum profit and number of candies. Is the maximum profit represented by the x-value or the y-value?
Please help! I know p(x)=r(x)-c(x), however i am having difficulty executing the function to find maximum profit..
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! tricky problem that's very difficult for somebody who hasn't been exposed to the procedure normally used in problems of this type.
revenue equation might look something like this:
y = (P + px) * (D - dx).
P would be the normal price per unit
D would be the normal units of demand.
p would be the change in the price per unit for each increment of change.
d would be the change in the units of demand for each increment of change.
x would represent the increments of change.
cost equation might look something like this:
y = (C + cx) * (D - dx).
C would the the normal cost per unit.
D would be the normal units of demand.
c would be the change in the cost per unit for each increment of change..
d would be the change in the units of demand for each increment of change.
based on this revenue and cost equation, the profit equation profit = revenue minus cost and would look something like this:
y = (P + px) * (D - dx) - (C + cx) * (D - dx).
y would represent the profit.
(P + px) * (D - dx) would represent the revenue.
(C + cx) * (D - dx) would represent the cost.
an example might look something like this:
assume the normal price per box of candy is 5 dollars.
assume the normal cost per box of candy is 2 dollars.
assume that the normal demand is 100,000 boxes of candy.
assume that the price per unit goes up 50 cents a box for each additional increment of change.
assume that the demand goes down 1,000 units for each additional increment of change.
this means that, every time the price goes up 50 cents, the demand goes down 1,000 boxes.
assume that the cost per unit goes up 10 cents for each additional increment of change.
this means that every time the demand goes down 1,000 units, the cost per box goes up 10 cents.
your general profit equation is:
y = (P + px) * (D - dx) - (C + cx) * (D - dx).
when P = 5 dollars and C = 2 dollars and D = 100,000 and p = 50 cents and c = 10 cents and d = 1,000, then the profit equation becomes:
y = (5 +.5 * x) * (100,000 - 1,000 * x) - (2 + .1 * x) * (100,000 - 1,000 * x)
to find the maximum point in this equation, you can put the equation in standard quadratic form and solve for the maximum value algebraically, or you can graph the equation and have the graph tell you what the maximum value for y is.
since the graph is easier, i'll do that first.
the graph will look like this:
you can see from the graph that the maximum profit occurs when x = 46.25.
when x = 46.25, the profit is shown as 1,155,625 dollars.
to understand what this mean, we go back to the equation used in the graph.
that equation is:
y = (5 + .5x) * (100,000 - 1,000 * x) - (2 + .1 * x) * (100,000 - 1,000 * x)
when x is 0, the equation becomes:
y = 5 * 100,000 - 2 * 100,000.
that gives a profit of 500,000 - 200,000 = 300,000
when x = 46.25, the equation becomes:
y = (5 + .5 * 46.25) * (100,000 - 1,000 * 46.25) - (2 + .1 * 46.25) * (100,000 - 1,000 * 46.25)
simplify this equation to get:
y = 28.125 * 53,750) - 6.625 * 53,750.
28.125 is the price after 46.25 increments of change are added.
53,750 is the demand after 46.25 increments of change are added.
simplify further to get:
y = 1,511,718.75 - 356,093.75 = 1,155,625.
maximum profit is 1,155,625 dollars when they charge 28.125 per box and sell 53,750 boxes of candy.
the revenue had become 1,511,718.75 and the cost had become 356,093.75.
the profit is the revenue minus the cost which is equal to 1,155,625 dollars.
since the graphing software i used told me what the maximum point on the graph was, it was easy to obtain it.
to solve this algebraically, you would convert the equation into standard quadratic form.
start with:
y = (5 + .5 * x) * (100,000 - 1,000 * x) - (2 + .1 * x) * (100,000 - 1,000 * x)
(5 + .5 * x) * (100,000 - 1,000 * x) becomes:
500,000 - 5,000 * x + 50,000 * x - 500 * x^2
combine like terms to get:
500,000 + 45,000 * x - 500 * x^2.
reorder the terms in descending order of degree to get:
-500 * x^2 + 45,000 * x + 500,000
this is the revenue part of the equation.
(2 + .1 * x) * (100,000 - 1,000 * x) becomes:
200,000 - 2,000 * x + 10,000 * x - 100 * x^2.
combine like terms to get:
200,000 + 8,000 * x - 100 * x^2.
reorder the terms in descending order of degree to get:
-100 * x^2 + 8,000 * x + 200,000.
this is the cost pert of the equation.
you now want to subtract the cost from the revenue to get the profit.
that would be:
-500 * x^2 + 45,000 * x + 500,000 - (-100 * x^2 + 8,000 * x - 100 * x^2.
simplify this to get:
-500 * x^2 + 45,000 * x + 500,000 + 100 * x^2 - 8,000 * x - 200,000.
combine like terms to get:
y = -400 * x^2 + 37,000 * x + 300,000.
this is your profit equation in standard quadratic form.
this equation is identical to the original profit equation of:
y = (5 + .5 * x) * (100,000 - 1,000 * x) - (2 + .1 * x) * (100,000 - 1,000 * x)
if you place both equations on the same graph, the graph will be identical, meaning that the same graph applies to both equation.
that is seen below, where i placed both equations on the same graph.
you can see that there is only one graph.
that means both equations have generated the same graph, and are therefore identical to each other.
now that the equation is in standard form, you can solve for the maximum value algebraically.
when you set y = 0, the standard form of the equation becomes
-400 * x^2 + 37,000 * x + 300,000 = 0
in this form, a is the coefficient of the x^2 term, b is the coefficient of the x term, c is the constant term.
you get a = -400, b = 27,000, c = 300,000.
the maximum value of this equation will be at x = -b/2a.
x = -b/2a bcomes x = -37,000 / -800.
this results in x = 46.25
you then find the value of y by replacing x in the equation with 46.25.
the equation becomes:
y = -400 * (46.25)^2 + 37,000 * 46.25 + 300,000.
solve for y to get y = 1,155,625.
this agrees with the graphical solution.
you get the same solution, whether you solve it graphically, or whether you solve it algebraically, as you should.
in summary:
you were correct in assuming that profit = revenue - cost and that, therefore, p(x) = r(x) - c(x)
when we made r(x) = (P + px) * (D - dx), and we made c(x) = (C + cx) * (D - dx), we got p(x), which became: p(x) = r(x) - c(x), which became:
p(x) = (P + px) * (D - dx) - (C + cx) * (D - dx).
the equation assumes that there is a normal price and a normal cost and a normal demand that changes by a specific amount with each change in the value of x.
i hope this helps.
it's a difficult question to answer simply, but i tried to make it as simple as possible, without making it too simple.
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