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January 30, 2015

January 30, 2015

Posted by **anonymous** on Thursday, February 10, 2011 at 11:03pm.

a) what is the largest number of cases she can sell and still make a point?

b) Explain how it is possible for her to lose money if she sells more cases than your answer in part (a)

I've been doing this problem since lastnite and i couldn;t figure out the answer. someone help me please? ireally need to learn how to solve this kind of problem. midterm is coming :[

- MAth -
**Reiny**, Thursday, February 10, 2011 at 11:15pmI don't know what level of math you are taking, is it Calculus?

if you find the vertex (by Calculus or completing the square), you should have had (15, 330)

which means that if she sells 15 cases she makes a maximum profit of 330.

Your function is a parabola which opens downwards, so any other value besides x=15 would produce a smaller value than 330

this parabola crosses the x-axis at appr. 2.1 and 27.8

so it is positive for number of cases between 3 and 27 cases.

For any other number of cases, P(x) would be negative.

e.g. let x = 30 cases

P(30) = -120

- MAth -
**anonymous**, Thursday, February 10, 2011 at 11:26pmthanks for the help really appreciate it! im taking business math. anyway, how did you find out that the parabola crosses the x-axis at appro. 2.1 and 27.8? and how did you come up with 27 cases?

sorry for asking a lotta questions. i just want to understand this problem.

Thanks again

- MAth -
**Reiny**, Friday, February 11, 2011 at 7:29amSorry about not answering last night, but had gone to bed.

I solved

0 = -2x^2+60x-120 using the quadratic formula

since x has to be a whole number, 3 and 27 would be the first and last whole numbers for which the parabola is above the x-axis, that is , for which P is positive.

Try subbing in x = 28 and x = 2, you will get a negative value for P.

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