Question

The profit for a product can be described by the function P(x) = 202x-5500 - x? dollars, where x is the number of units produced and sold. To maximize profit, howmany units must be produced and sold? What is the maximum possible profit?To maximize profit, units must be produced and sold.(Simplify your answer.)The maximum possible profit is so(Simplify your answer.)

Answer 1

Given that the profit for a product is described as:

`P(x)=202x-5500-x^2`

1. Number of units produced for maximum profit.

To determine the number of units produced for maximum profit,

step 1: Differentiate the profit function.

`\begin{gathered} P(x)=202x-5500-x^2 \\ P^(\prime)(x)=(dP(x))/(dx)=202-2x \end{gathered}`

step 2: Determine the turning point of the function.

At the turning point, P'(x) equals zero.

Thus,

`\begin{gathered} P^(\prime)(x)=(dP(x))/(dx)=202-2x \\ P^(\prime)(x)=0 \\ \text{thus,} \\ 202-2x=0 \\ \text{subtract 202 from both sides of dthe equation. thus,} \\ 202-202-2x=0-202 \\ -2x=-202 \\ \text{divide both sides of the equation by the coefficient of x, which is -2.} \\ (-2x)/(-2)=-(202)/(-2) \\ \Rightarrow x=101 \end{gathered}`

step 3: Determine the extreme point(s) of the function by evaluating the second differential of the function.

`\begin{gathered} \text{when }P^(\doubleprime)(x)<0,\text{ the function has a ma}ximum\text{ point.} \\ \text{when }P^(\doubleprime)(x)>0,\text{ the function has a minimum point.} \end{gathered}`

thus, the second differential is evaluated as

`P^(\doubleprime)(x)=(d^2P(x))/(dx^2)=-2`

Since, the second differential is negative as shown above, the profit function has a maximum.

Thus, the number of units to be produced for maximum profit is 101 units.

2. Maximum profit

To evaluate the maximum profit, substitute the number of units produced which brings about maximum profit into the profit function.

Thus,

`\begin{gathered} P(x)=202x-5500-x^2 \\ \text{where x=101} \\ \text{thus,} \\ P(x)_(\max )=202(101)-5500-(101)^2 \\ =4701 \end{gathered}`

Hence, the maximum possible profit is 4701 dollars.