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?
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
a) To determine if April could ever earn a profit of $2000, we can substitute P = 2000 into the profit model equation and see if there is a solution.
-0.1n^2 + 30n - 1200 = 2000
Rearranging the equation, we get:
-0.1n^2 + 30n - 3200 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
n = [-b ± √(b^2 - 4ac)] / 2a
In this case, a = -0.1, b = 30, and c = -3200. Plugging in these values, we get:
n = [-30 ± √(30^2 - 4(-0.1)(-3200))] / 2(-0.1)
Simplifying further, we have:
n = [-30 ± √(900 + 1280)] / (-0.2)
n = [-30 ± √2180] / (-0.2)
Calculating the square root of 2180, we get:
n ≈ [-30 ± 46.67] / (-0.2)
This yields two solutions:
n ≈ (-30 + 46.67) / (-0.2) ≈ 83.33
n ≈ (-30 - 46.67) / (-0.2) ≈ 238.33
Since the number of teddy bears sold cannot be fractional or negative, April cannot earn a profit of $2000 based on this model.
b) To break even, April's profit should be zero. We need to solve the equation P = 0.
-0.1n^2 + 30n - 1200 = 0
We can use the quadratic formula again:
n = [-b ± √(b^2 - 4ac)] / 2a
Plugging in the values, we have:
n = [-30 ± √(30^2 - 4(-0.1)(-1200))] / 2(-0.1)
Simplifying further, we get:
n = [-30 ± √(900 - 480)] / (-0.2)
n = [-30 ± √420] / (-0.2)
Calculating the square root of 420, we have:
n ≈ [-30 ± 20.49] / (-0.2)
This yields two solutions:
n ≈ (-30 + 20.49) / (-0.2) ≈ 96.77
n ≈ (-30 - 20.49) / (-0.2) ≈ 254.92
Since the number of teddy bears sold cannot be fractional or negative, April would need to sell approximately 97 teddy bears to break even.
c) To earn $500, we need to solve the equation P = 500.
-0.1n^2 + 30n - 1200 = 500
Rearranging the equation, we get:
-0.1n^2 + 30n - 1700 = 0
We can use the quadratic formula again:
n = [-b ± √(b^2 - 4ac)] / 2a
Plugging in the values, we have:
n = [-30 ± √(30^2 - 4(-0.1)(-1700))] / 2(-0.1)
Simplifying further, we get:
n = [-30 ± √(900 + 680)] / (-0.2)
n = [-30 ± √(1580)] / (-0.2)
Calculating the square root of 1580, we have:
n ≈ [-30 ± 39.75] / (-0.2)
This yields two solutions:
n ≈ (-30 + 39.75) / (-0.2) ≈ 148.75
n ≈ (-30 - 39.75) / (-0.2) ≈ 348.75
Since the number of teddy bears sold cannot be fractional or negative, April would need to sell approximately 149 teddy bears to earn $500.
d) To maximize her profit, we can use the vertex formula:
n = -b / (2a)
In this case, a = -0.1 and b = 30. Plugging these values into the formula, we have:
n = -30 / (2 * -0.1) = -30 / (-0.2) = 150
Therefore, April would need to sell 150 teddy bears to maximize her profit.
To answer these questions, we will use the profit function given by P = -0.1n^2 + 30n - 1200, where P represents the profit in dollars and n represents the number of teddy bears sold.
a.) To determine if April could ever earn a profit of $2000, we can substitute P = 2000 into the profit function and solve for n:
2000 = -0.1n^2 + 30n - 1200
Rearranging the equation, we get:
0.1n^2 - 30n + 3200 = 0
We can solve this quadratic equation by factoring or by using the quadratic formula. Let's use the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 0.1, b = -30, and c = 3200. Plugging these values into the formula, we get two possible solutions for n. Check if any of these solutions are viable values for n.
b.) To find the number of teddy bears April needs to sell to break even, we need to determine when her profit is zero. This means we need to solve the equation P = 0:
-0.1n^2 + 30n - 1200 = 0
Again, we can solve this quadratic equation using factoring or the quadratic formula. Once we find the values for n, we can determine the number of teddy bears April needs to sell to break even.
c.) To find the number of teddy bears April needs to sell to earn $500, we need to determine when her profit is $500. This means we need to solve the equation P = 500:
-0.1n^2 + 30n - 1200 = 500
Similarly, solving this quadratic equation will give us the values for n, which represent the number of teddy bears April needs to sell to earn $500.
d.) To determine the number of teddy bears April needs to sell to maximize her profit, we need to find the vertex of the parabolic function represented by the profit equation. The x-coordinate of the vertex will give us the number of teddy bears she needs to sell.
The x-coordinate of the vertex is given by -b / (2a), where a = 0.1 and b = 30. Compute this value to find the number of teddy bears needed for maximum profit.
By solving these equations and using the quadratic formula, we can find the values for n that answer each of these questions.