math

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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 -

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