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
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