Question

the profit for a product can be described by the function ( p(x) = 202x - 5000 - x^{2} ) 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? to maximize profit, units must be produced and sold. (simplify your answer)

Explanation:

Step1: Find the derivative of the profit function

The profit function is \(P(x)=202x - 5000-x^{2}\).
Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(P^\prime(x)=(202x)^\prime-(5000)^\prime-(x^{2})^\prime\).
Since \((ax)^\prime=a\) (where \(a = 202\)) and \((c)^\prime=0\) (where \(c = 5000\)), we have \(P^\prime(x)=202-0 - 2x=202-2x\).

Step2: Set the derivative equal to zero and solve for \(x\)

To find the critical points, set \(P^\prime(x)=0\).

$$
LATEXBLOCK0
$$

Step3: Check the second - derivative

The second - derivative \(P^{\prime\prime}(x)=(202 - 2x)^\prime=-2\).
Since \(P^{\prime\prime}(x)=-2<0\), the function \(P(x)\) is concave down at \(x = 101\), and \(x = 101\) is a point of maximum.

Answer:

\(101\)