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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): 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.
| x | P(x) | | 2113 | 5017.3003 | | 2114 | 5017.3052 | | 2115 | 5017.3075 | | 2116 | 5017.3072 | | 2117 | 5017.3043 |
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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