The profit for a product can be described by the function ​P(x)equals202xminus5500minusxsquared ​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?

Question

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Part 1
To maximize​ profit,
  
enter your response here units must be produced and sold.
​(Simplify your​ answer.)
Part 2
The maximum possible profit is ​$
  
enter your response here.
​(Simplify your​ answer.)

Answer 1

To find the number of units that must be produced and sold to maximize profit, we start with the given profit function:

\[ P(x) = 202x - 5500 - x^2 \]

This is a quadratic function in the form \(P(x) = ax^2 + bx + c\). Here, \(a = -1\), \(b = 202\), and \(c = -5500\). For a parabola that opens downwards (since \(a < 0\)), the maximum profit occurs at the vertex of the parabola.

The x-coordinate of the vertex can be found using the formula:

\[ x = -\frac{b}{2a} \]

Substituting in the values for \(a\) and \(b\):

\[ x = -\frac{202}{2 \cdot (-1)} = \frac{202}{2} = 101 \]

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

Next, we find the maximum possible profit by substituting \(x = 101\) back into the profit function:

\[ P(101) = 202(101) - 5500 - (101)^2 \]

Calculating each term:

1. \(202 \times 101 = 20402\)
2. \((101)^2 = 10201\)

Now plug these into the profit equation:

\[ P(101) = 20402 - 5500 - 10201 \] \[ P(101) = 20402 - 5500 - 10201 = 20402 - 15701 = 4701 \]

Thus, the maximum possible profit is $4701.

Summary of Answers:

• To maximize profit, 101 units must be produced and sold.
• The maximum possible profit is $4701.