Question
A company can produce a maximum of 1500 widgets in a year. If they sell x widgets during the year then their profit, in dollars, is given by, P(x)=30,000,000-360,000x+750x^2-(1/3)x^3 How many widgets should they try to sell in order to maximize their profit? Why are there 2 answers? What is the amount of profit with the 2 amounts of widgets? Which is the true profit then? Is the value fpr x, number of widgets possible for the company? Why or why not? What is the marginal profit function? What is the marginal profit if x=1000?
Answers
GPT-4o mini
To find the number of widgets that maximizes profit for the given profit function P(x):
\[
P(x) = 30,000,000 - 360,000x + 750x^2 - \frac{1}{3}x^3
\]
1. **Find the critical points**: Differentiate the profit function with respect to \(x\) and set it to zero.
\[
P'(x) = -360,000 + 1500x - x^2
\]
Setting the derivative to zero, we have:
\[
-360,000 + 1500x - x^2 = 0
\]
Rearranging gives:
\[
x^2 - 1500x + 360,000 = 0
\]
Now, we can apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 1\), \(b = -1500\), and \(c = 360,000\):
\[
x = \frac{1500 \pm \sqrt{(-1500)^2 - 4(1)(360,000)}}{2(1)} = \frac{1500 \pm \sqrt{2250000 - 1440000}}{2} = \frac{1500 \pm \sqrt{810000}}{2}
\]
Calculating \(\sqrt{810000}\):
\[
\sqrt{810000} = 900
\]
Thus, we have:
\[
x = \frac{1500 \pm 900}{2}
\]
Solving this gives two potential solutions:
\[
x = \frac{2400}{2} = 1200
\]
\[
x = \frac{600}{2} = 300
\]
2. **Profit with both amounts of widgets**:
For \(x = 1200\):
\[
P(1200) = 30,000,000 - 360,000(1200) + 750(1200)^2 - \frac{1}{3}(1200)^3
\]
Calculating step by step:
- \(360,000 \times 1200 = 432,000,000\)
- \(750 \times 1200^2 = 750 \times 1,440,000 = 1,080,000,000\)
- \(\frac{1}{3} \times 1200^3 = \frac{1}{3} \times 1,728,000,000 = 576,000,000\)
Putting it all together:
\[
P(1200) = 30,000,000 - 432,000,000 + 1,080,000,000 - 576,000,000 = 30,000,000 + 72,000,000 = 102,000,000
\]
For \(x = 300\):
\[
P(300) = 30,000,000 - 360,000(300) + 750(300)^2 - \frac{1}{3}(300)^3
\]
Calculating step by step:
- \(360,000 \times 300 = 108,000,000\)
- \(750 \times 300^2 = 750 \times 90,000 = 67,500,000\)
- \(\frac{1}{3} \times 300^3 = \frac{1}{3} \times 27,000,000 = 9,000,000\)
Putting it all together:
\[
P(300) = 30,000,000 - 108,000,000 + 67,500,000 - 9,000,000 = 30,000,000 - 49,500,000 + 67,500,000 = 48,000,000
\]
So, the profit for \(x = 1200\) is \(102,000,000\) and profit for \(x = 300\) is \(48,000,000\).
3. **Determining the true profit**: The maximum profit occurs at \(x = 1200\), as it yields a higher profit than \(x = 300\).
4. **Checking feasibility**: Both \(x = 300\) and \(x = 1200\) are feasible production levels since they are within the maximum production capacity of the company (1500 widgets).
5. **Marginal profit function**: The marginal profit function \( P'(x) \):
\[
P'(x) = -360,000 + 1500x - x^2
\]
6. **Marginal profit at \(x = 1000\)**:
\[
P'(1000) = -360,000 + 1500(1000) - (1000)^2
\]
Calculating:
\[
P'(1000) = -360,000 + 1,500,000 - 1,000,000 = 140,000
\]
Thus, the marginal profit when producing 1000 widgets is \(140,000\) dollars.
\[
P(x) = 30,000,000 - 360,000x + 750x^2 - \frac{1}{3}x^3
\]
1. **Find the critical points**: Differentiate the profit function with respect to \(x\) and set it to zero.
\[
P'(x) = -360,000 + 1500x - x^2
\]
Setting the derivative to zero, we have:
\[
-360,000 + 1500x - x^2 = 0
\]
Rearranging gives:
\[
x^2 - 1500x + 360,000 = 0
\]
Now, we can apply the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \(a = 1\), \(b = -1500\), and \(c = 360,000\):
\[
x = \frac{1500 \pm \sqrt{(-1500)^2 - 4(1)(360,000)}}{2(1)} = \frac{1500 \pm \sqrt{2250000 - 1440000}}{2} = \frac{1500 \pm \sqrt{810000}}{2}
\]
Calculating \(\sqrt{810000}\):
\[
\sqrt{810000} = 900
\]
Thus, we have:
\[
x = \frac{1500 \pm 900}{2}
\]
Solving this gives two potential solutions:
\[
x = \frac{2400}{2} = 1200
\]
\[
x = \frac{600}{2} = 300
\]
2. **Profit with both amounts of widgets**:
For \(x = 1200\):
\[
P(1200) = 30,000,000 - 360,000(1200) + 750(1200)^2 - \frac{1}{3}(1200)^3
\]
Calculating step by step:
- \(360,000 \times 1200 = 432,000,000\)
- \(750 \times 1200^2 = 750 \times 1,440,000 = 1,080,000,000\)
- \(\frac{1}{3} \times 1200^3 = \frac{1}{3} \times 1,728,000,000 = 576,000,000\)
Putting it all together:
\[
P(1200) = 30,000,000 - 432,000,000 + 1,080,000,000 - 576,000,000 = 30,000,000 + 72,000,000 = 102,000,000
\]
For \(x = 300\):
\[
P(300) = 30,000,000 - 360,000(300) + 750(300)^2 - \frac{1}{3}(300)^3
\]
Calculating step by step:
- \(360,000 \times 300 = 108,000,000\)
- \(750 \times 300^2 = 750 \times 90,000 = 67,500,000\)
- \(\frac{1}{3} \times 300^3 = \frac{1}{3} \times 27,000,000 = 9,000,000\)
Putting it all together:
\[
P(300) = 30,000,000 - 108,000,000 + 67,500,000 - 9,000,000 = 30,000,000 - 49,500,000 + 67,500,000 = 48,000,000
\]
So, the profit for \(x = 1200\) is \(102,000,000\) and profit for \(x = 300\) is \(48,000,000\).
3. **Determining the true profit**: The maximum profit occurs at \(x = 1200\), as it yields a higher profit than \(x = 300\).
4. **Checking feasibility**: Both \(x = 300\) and \(x = 1200\) are feasible production levels since they are within the maximum production capacity of the company (1500 widgets).
5. **Marginal profit function**: The marginal profit function \( P'(x) \):
\[
P'(x) = -360,000 + 1500x - x^2
\]
6. **Marginal profit at \(x = 1000\)**:
\[
P'(1000) = -360,000 + 1500(1000) - (1000)^2
\]
Calculating:
\[
P'(1000) = -360,000 + 1,500,000 - 1,000,000 = 140,000
\]
Thus, the marginal profit when producing 1000 widgets is \(140,000\) dollars.