Your company would like to know how sales levels affect profits. If too few items are sold, then there is a loss. Even if too many items are sold, however, the company can lose money (likely because of low pricing). It is good to know how many items can be sold for there to be profit.
P(x) = –x2 + 110x – 1,000
This function can be used to compute the profit (in thousands of dollars) from producing and selling a certain number, x, of thousands of smartphones.
Compute the following: P(5), P(50), and P(120). Then, interpret the results.
Discuss and interpret the meaning where the profit function crosses the x-axis. Refer to last week’s assignment concerning break-even points, and interpret the graph. Also, discuss where the graph is above and below the x-axis, explaining what that means in terms of profitability.
( I am completely lost here I don't even know where to start please help)
SUBSTITUTE THE VALUE IN THE PARENTHESES FOR X - 3 TIMES FOR 5, 50 AND 120
To compute the values of P(5), P(50), and P(120), you need to substitute the values of x into the profit function P(x) = –x^2 + 110x – 1,000 and calculate the result.
Let's start with P(5):
P(5) = –5^2 + 110(5) – 1,000
= –25 + 550 – 1,000
= –475
Next, let's calculate P(50):
P(50) = –50^2 + 110(50) – 1,000
= –2,500 + 5,500 – 1,000
= 2,000
Finally, let's compute P(120):
P(120) = –120^2 + 110(120) – 1,000
= –14,400 + 13,200 – 1,000
= –2,200
Now let's interpret the results:
1. P(5) = -475: This means that if the company produces and sells 5,000 smartphones, the profit will be a loss of $475,000. This indicates that the company is selling too few items, resulting in a loss.
2. P(50) = 2,000: This shows that if the company produces and sells 50,000 smartphones, the profit will be $2,000,000. This suggests that the company has found the optimal quantity to maximize profits.
3. P(120) = -2,200: For the sale of 120,000 smartphones, the profit will be a loss of $2,200,000. This indicates that the company is selling too many items, resulting in a loss due to potentially lower pricing.
Now let's consider where the profit function crosses the x-axis (when P(x) = 0) and its implications:
To find where P(x) = 0, we set the equation equal to zero and solve for x:
–x^2 + 110x – 1,000 = 0
By solving this quadratic equation, you can find the x-values at which the profit is zero, known as the break-even points. In this case, there may be two break-even points, assuming there are real solutions.
When P(x) > 0: This means that the profit is positive, indicating a profit is being made. The company is earning more revenue than the costs involved in production and sales.
When P(x) < 0: This suggests that the profit is negative, resulting in a loss. The company is not making enough revenue to cover the costs involved, indicating potential issues with pricing or expenses.
In terms of profitability, when the graph of the profit function is above the x-axis, it means that the company is making a profit. When the graph is below the x-axis, it signifies that the company is incurring losses.
Interpreting the graph will provide more insights into the relationship between sales levels and profitability.