Posted by **geekgirl95** on Monday, June 20, 2011 at 10:15pm.

april sells specialty teddy bears at various summer festivals. her profit for a week,P, in dollars, can be modelled by P= -0.1n^2 + 30n - 1200, where n is the umber of teddy bearsshe sells during the week.

a.) According to this model, could april ever earn a profit of $2000.

b.) how many teddy bears would she have to sell to break even ?

c.) how many teddy bears to earn $500?

d.) how many teddy bears would she have to sell to maximize her profit?

- math -
**MathMate**, Tuesday, June 21, 2011 at 9:44am
Given Profit, P = -0.1n^2 + 30n - 1200

a.

If P=2000, then

2000=-0.1n^2 + 30n - 1200

Solve for n in

0.1n^2 - 30n + 1200 + 2000 = 0

or

n^2 - 300n + 32000 = 0

Since the discriminant (-300)²-4*3200 < 0

there are no real roots for n, hence April will not be able to get a profit of $2000.

b. to break even means P=0

c. to make a profit of $500 means P=500

d. to maximize profit will require a little trial and error (it is between n=100 and n=200).

However, if you have learned calculus, you can equate P'(n)=0. If not, you can also evaluate the value of n to give a Pmax at n=-b/(2a) in the equation

P(n)=a*n^2 + b*n + c

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