SOLUTION: A candy store finds that it can make a profit of P dollars each month by selling x boxes of candy. Using the formula: P(x)=-.0013x^2+5.5x-800,how many boxes of candy must the store

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Question 1208038: A candy store finds that it can make a profit of P dollars each month by selling x boxes of candy. Using the formula: P(x)=-.0013x^2+5.5x-800,how many boxes of candy must the store sell in order to maximize their profits?what is the maximum profit?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52878)   (Show Source): You can put this solution on YOUR website!
.
A candy store finds that it can make a profit of P dollars each month by selling x boxes of candy.
Using the formula: P(x)=-.0013x^2+5.5x-800, how many boxes of candy must the store sell
in order to maximize their profits? what is the maximum profit?
~~~~~~~~~~~~~~~~~~~ 

In this problem, the profit is a quadratic function representing a parabola.


Since the coefficient at x^2 is negative (it is -0.0013), the parabola is opened downward and has the maximum.


The maximum of any quadratic function y = ax^2 + bx + c with negative coefficient  "a"  is at

     = .


In our case it is   =  =  = 2115.384615.


Since the number of boxes is an integer number, we should round the number to 2115.


The maximum profit is then   =  = 5017.31 dollars (rounded).

At this point, the problem is solved in full.

-------------------

On finding the maximum/minimum of a quadratic function see the lessons

    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola

Consider these lessons as your textbook,  handbook,  tutorials and  (free of charge)  home teacher,
who is always with you to help.

Learn the subject from there once and for all.



Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Compare P(x) = -0.0013x^2+5.5x-800 with the quadratic template P(x) = ax^2+bx+c
a = -0.0013
b = 5.5
c = -800

The max profit occurs at the vertex (h,k) which is the highest point on the parabola (only when a < 0)

Let's find the x coordinate of the vertex.
h = -b/(2*a)
h = (-5.5)/(2*(-0.0013))
h = 2115.384615 approximately
Of course it's not possible to sell a fractional amount of candy boxes.
Let's try out integer values close to 2115.
We'll try values of x from the set {2113, 2114, 2115, 2116, 2117}

I'll leave the scratch work for the student to do, but you should get this table of values.
xP(x)
21135017.3003
21145017.3052
21155017.3075
21165017.3072
21175017.3043

I recommend using a spreadsheet.

The 2nd column of values are exact and haven't been rounded.
When rounding to the nearest penny, the first and last results round to 5017.30; while the middle 3 items round to 5017.31 which is the max profit possible.

So you could sell either 2114, 2115, or 2116 boxes of candy to get the same max profit.
Usually with these types of problems there's only one possible x value that leads to the max P(x) value. So it's a bit interesting we get three x values instead.

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Summary:

How many boxes to sell? Either 2114, 2115, or 2116 boxes

Max profit: $5,017.31
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