Question
A small company shows the profits from their business with the function P(x)equals=minus−0.01xsquared2plus+60xminus−500, where x is the number of units they sell and P is the profit in dollars.
a. How many units are sold by the company to earn the maximum profit?
b. Between which numbers of units sold does the company show a profit?
a. How many units are sold by the company to earn the maximum profit?
b. Between which numbers of units sold does the company show a profit?
Answers
GPT-4o mini
To solve the problem, we have the profit function given by:
\[
P(x) = -0.01x^2 + 60x - 500
\]
This is a quadratic function in the standard form \(P(x) = ax^2 + bx + c\), where:
- \(a = -0.01\) (the coefficient of \(x^2\))
- \(b = 60\) (the coefficient of \(x\))
- \(c = -500\) (the constant term)
### a. Finding the number of units sold to earn maximum profit
The maximum profit for a quadratic function occurs at the vertex. The x-coordinate of the vertex can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Substituting in our values:
\[
x = -\frac{60}{2 \cdot -0.01} = -\frac{60}{-0.02} = 3000
\]
So, the company needs to sell **3000 units** to earn the maximum profit.
### b. Finding the range of units sold where the company shows a profit
The company shows a profit when \(P(x) > 0\). We can find the values of \(x\) that make \(P(x) = 0\) by solving the equation:
\[
-0.01x^2 + 60x - 500 = 0
\]
We can multiply through by -100 to eliminate the decimal:
\[
x^2 - 6000x + 50000 = 0
\]
Now, use the quadratic formula, where \(a = 1\), \(b = -6000\), and \(c = 50000\):
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Calculating the discriminant:
\[
b^2 - 4ac = (-6000)^2 - 4 \cdot 1 \cdot 50000 = 36000000 - 200000 = 35800000
\]
Now we can find \(x\):
\[
x = \frac{6000 \pm \sqrt{35800000}}{2 \cdot 1}
\]
Calculating \(\sqrt{35800000}\):
\[
\sqrt{35800000} \approx 5973.09
\]
Now substituting back in:
\[
x \approx \frac{6000 \pm 5973.09}{2}
\]
Calculating the two possible values for \(x\):
1. \(x_1 = \frac{6000 + 5973.09}{2} \approx \frac{11973.09}{2} \approx 5986.54\)
2. \(x_2 = \frac{6000 - 5973.09}{2} \approx \frac{26.91}{2} \approx 13.455\)
Thus, the company shows a profit between the values:
\[
13.455 < x < 5986.54
\]
Since \(x\) represents the number of units sold, we round to the nearest whole units. Therefore, the company shows a profit when:
\[
\text{between } 14 \text{ and } 5986 \text{ units sold.}
\]
### Summary of Results:
**a.** Maximum profit is achieved by selling **3000 units**.
**b.** The company shows a profit between **14 and 5986** units sold.
\[
P(x) = -0.01x^2 + 60x - 500
\]
This is a quadratic function in the standard form \(P(x) = ax^2 + bx + c\), where:
- \(a = -0.01\) (the coefficient of \(x^2\))
- \(b = 60\) (the coefficient of \(x\))
- \(c = -500\) (the constant term)
### a. Finding the number of units sold to earn maximum profit
The maximum profit for a quadratic function occurs at the vertex. The x-coordinate of the vertex can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Substituting in our values:
\[
x = -\frac{60}{2 \cdot -0.01} = -\frac{60}{-0.02} = 3000
\]
So, the company needs to sell **3000 units** to earn the maximum profit.
### b. Finding the range of units sold where the company shows a profit
The company shows a profit when \(P(x) > 0\). We can find the values of \(x\) that make \(P(x) = 0\) by solving the equation:
\[
-0.01x^2 + 60x - 500 = 0
\]
We can multiply through by -100 to eliminate the decimal:
\[
x^2 - 6000x + 50000 = 0
\]
Now, use the quadratic formula, where \(a = 1\), \(b = -6000\), and \(c = 50000\):
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Calculating the discriminant:
\[
b^2 - 4ac = (-6000)^2 - 4 \cdot 1 \cdot 50000 = 36000000 - 200000 = 35800000
\]
Now we can find \(x\):
\[
x = \frac{6000 \pm \sqrt{35800000}}{2 \cdot 1}
\]
Calculating \(\sqrt{35800000}\):
\[
\sqrt{35800000} \approx 5973.09
\]
Now substituting back in:
\[
x \approx \frac{6000 \pm 5973.09}{2}
\]
Calculating the two possible values for \(x\):
1. \(x_1 = \frac{6000 + 5973.09}{2} \approx \frac{11973.09}{2} \approx 5986.54\)
2. \(x_2 = \frac{6000 - 5973.09}{2} \approx \frac{26.91}{2} \approx 13.455\)
Thus, the company shows a profit between the values:
\[
13.455 < x < 5986.54
\]
Since \(x\) represents the number of units sold, we round to the nearest whole units. Therefore, the company shows a profit when:
\[
\text{between } 14 \text{ and } 5986 \text{ units sold.}
\]
### Summary of Results:
**a.** Maximum profit is achieved by selling **3000 units**.
**b.** The company shows a profit between **14 and 5986** units sold.