A stereo manufacturer determines that in order to sell x units of a new stereo, the price per unit , in dollars, must be p(x) = 100 - x. The manufacturer also determines that the total cost of producing x units is given by C(x) = 3000 + 2x. a. Express the total revenue function R as a function of x. b. Express the total profit P as a function of x. c. How many units must the company produce and sell in order to maximize profit d. What is the maximum profit? e. What price per unit must be charged in order to make this maximum profit .

[tex]$The total revenue R(x) is obtained as:\begin{align*} R\left( x \right) &= xp\left( x \right)\\ &= x\left( {100 - x} \right)\\ &= 100x - {x^2} &$$ R(x) = 100x - {x^2} \end{align*}[/tex]

[tex]$Thus, {R\left( x \right) = 100x - {x^2}[/tex]

b)

[tex]$The total profit is obtained as: \begin{align*} P\left( x \right) &= R\left( x \right) - C\left( x \right)\\ &= 1000x - {x^2} - 3000 - 20x\\ &= - {x^2} + 980x - 3000 \end{align*}[/tex]

[tex]$Hence, P\left( x \right) = - {x^2} + 980x - 3000[/tex]

c)

[tex]$For maximum profit dP/dx should be equal to zero: \begin{align*} \dfrac{{dP}}{{dx}} &= 0\\ - 2x + 980 &= 0\\ x &= 490 \end{align*}[/tex]

[tex]$So, the required units to maximize profit is 490[/tex]

d)

[tex]$The maximum profit is calculated as: \begin{align*} P\left( x \right) &= - {x^2} + 980x - 3000\\ P\left( {490} \right) &= - {490^2} + 980 \times 490 - 3000\\ &= 237100 \end{align*}[/tex]

[tex]$Accordingly, the maximum profit is \$237100.[/tex]

e)

[tex]$Calculating the price at maximum profit as: \begin{align*} p\left( x \right) &= 1000 - x\\ p\left( {490} \right) &= 1000 - 490\\ &= 510 \end{align*}[/tex]

[tex]$Consequently, the price at the maximum profit is \$510.[/tex]