Question

The profit for a product can be described by the function ​P(x)equals202xminus5000minusxsquared ​dollars, where x is the number of units produced and sold. To maximize​ profit, how many units must be produced and​ sold? What is the maximum possible​ profit?
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Part 1
To maximize​ profit,
  
enter your response here units must be produced and sold.

Answer 1

To maximize the profit described by the function \( P(x) = 202x - 5000 - x^2 \), we can determine the number of units \( x \) that maximizes profit by using the vertex formula of a quadratic function.

The function can be rewritten as:
\[ P(x) = -x^2 + 202x - 5000 \]

This is a quadratic function of the form \( P(x) = ax^2 + bx + c \), where \( a = -1 \), \( b = 202 \), and \( c = -5000 \). For a quadratic function, the x-coordinate of the vertex (which gives us the maximum profit in this case since \( a < 0 \)) can be found using the formula:
\[ x = -\frac{b}{2a} \]

Substituting the values of \( a \) and \( b \):
\[ x = -\frac{202}{2(-1)} = \frac{202}{2} = 101 \]

Therefore, to maximize profit, **101 units** must be produced and sold.

Now we can calculate the maximum possible profit by substituting \( x = 101 \) back into the profit function:
\[ P(101) = 202(101) - 5000 - (101)^2 \]

Calculating each term:
- \( 202 \times 101 = 20402 \)
- \( (101)^2 = 10201 \)

Now substituting into the profit function:
\[ P(101) = 20402 - 5000 - 10201 \]
\[ P(101) = 20402 - 5000 - 10201 = 20402 - 15201 = 5201 \]

Thus, the maximum possible profit is **$5201**.

**Final answers:**
- Units to maximize profit: **101**
- Maximum profit: **$5201**