Question

The second-largest public utility in the nation is the sole provider of electricity in 32 counties of southern Florida. To meet the monthly demand for electricity in these counties, which is given by the inverse demand function P=1,200 - 4Q, the utility company has set up two electric generating facilities: Q1 kilowatts are produced at facility 1, and Q2 kilowatts are produced at facility 2 (so Q= Q1+ Q2). The costs of producing electricity at each facility are given by C1(Q1) = 8,000+6Q12 and C2(Q2) = 6,000 +3Q22, respectively. Determine the profit-maximizing amounts of electricity to produce at the two facilities, the optimal price, and the utility company’s profits.

Short answer

Answer

The profit-maximizing output at the facility 1 is 60 units and at the facility 2 is 420 units. The optimal price is $2020, and the company’s profit is $969,980.

Step-by-step solution

Marginal cost equation

Given:

An energy company in a monopolistically competitive market has the following demand and cost functions for the two facilities it produces:

Inverse demand function: $\mathit{P}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{4}\mathit{Q}$

Cost Function Facility 1: $\mathit{C}\mathbf{\left(}{\mathbf{Q}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{8}\mathbf{,}\mathbf{000}\mathbf{+}\mathbf{6}{\mathit{Q}}_{\mathbf{1}}^{\mathbf{2}}$

Cost Function Facility 1: $\mathit{C}\mathbf{\left(}{\mathbf{Q}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{6}\mathbf{,}\mathbf{000}\mathbf{+}\mathbf{3}{\mathit{Q}}_{\mathbf{2}}^{\mathbf{2}}$

In the monopolistic competition market, as in the monopoly, the same function is fulfilled, that production and profits are maximized when marginal income is equal (MR=MC). Therefore, the demand and cost functions are derived to obtain the marginal revenues and costs.

To obtain the marginal revenue, we derive total revenue (TR) first.

$$\begin{gathered} \mathit{T}\mathit{R}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{4}\mathbf{Q}\mathbf{)}\mathbf{\times }\mathit{Q} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{200}\mathit{Q}\mathbf{+}\mathbf{4}{\mathit{Q}}^{\mathbf{2}} \end{gathered}$$

Now the first derivative is applied to obtain the marginal income:

$\mathit{M}\mathit{R}\mathbf{=}\frac{\mathbf{dR}}{\mathbf{dQ}}\mathbf{=}\mathbf{1200}\mathbf{+}\mathbf{8}\mathit{Q}$

The rule tells us that $\mathit{Q}\mathbf{=}{\mathit{Q}}_{\mathbf{1}}\mathbf{+}{\mathit{Q}}_{\mathbf{2}}$so we can substitute the value for each level of production in the previous equation:

$$\begin{gathered} \mathit{M}\mathit{R}\mathbf{=}\mathbf{1200}\mathbf{+}\mathbf{8}\mathbf{(}{\mathbf{Q}}_{\mathbf{1}}\mathbf{+}{\mathbf{Q}}_{\mathbf{2}}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1200}\mathbf{+}\mathbf{8}\mathit{Q}\mathbf{1}\mathbf{+}\mathbf{8}{\mathit{Q}}_{\mathbf{2}} \end{gathered}$$

Marginal Cost Facility 1:

$$\begin{gathered} \mathit{C}\mathbf{\left(}{\mathbf{Q}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{8}\mathbf{,}\mathbf{000}\mathbf{+}\mathbf{6}{\mathit{Q}}_{\mathbf{1}}^{\mathbf{2}} \\ \mathit{M}\mathit{C}\mathbf{\left(}{\mathbf{Q}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{dC}}{\mathbf{dQ}}\mathbf{=}\mathbf{12}{\mathit{Q}}_{\mathbf{1}} \end{gathered}$$

Marginal Cost Facility 2:

$$\begin{gathered} \mathit{C}\mathbf{\left(}{\mathbf{Q}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}\mathbf{6}\mathbf{,}\mathbf{000}\mathbf{+}\mathbf{2}{\mathit{Q}}_{\mathbf{2}}^{\mathbf{2}} \\ \mathit{M}\mathit{C}\mathbf{\left(}{\mathbf{Q}}_{\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{dC}}{\mathbf{dQ}}\mathbf{=}\mathbf{4}{\mathit{Q}}_{\mathbf{2}} \end{gathered}$$

Given that we have two facilities and therefore two marginal costs, the relationship MR=MC must be applied for each facility to obtain the quantities of each one.

Function of maximizing

Function that maximizes quantity for facility 1:

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{M}\mathit{R}\mathbf{=}\mathit{M}\mathit{C}\left(Q_1\right) \\ \mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{8}{\mathit{Q}}_{\mathbf{1}}\mathbf{+}\mathbf{8}{\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{12}\mathit{Q} \end{gathered}$$

Equating to zero and adding equal terms:

$\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{-}\mathbf{4}{\mathit{Q}}_{\mathbf{1}}\mathbf{+}\mathbf{8}{\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

$$\begin{gathered} \mathbf{1}\mathbf{,}\mathbf{200}\mathbf{-}\mathbf{4}{\mathit{Q}}_{\mathbf{1}}\mathbf{+}\mathbf{8}{\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{M}\mathit{R}\mathbf{=}\mathit{M}\mathit{C}\mathbf{\left(}{\mathbf{Q}}_{\mathbf{2}}\mathbf{\right)} \end{gathered}$$

Equating to zero and adding equal terms:

$\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{8}{\mathit{Q}}_{\mathbf{1}}\mathbf{+}\mathbf{4}{\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

Having both functions of facility 1 and 2 that maximize production, we subtract and solve them to obtain the values of Q1and Q2:

$$\begin{gathered} \mathbf{1}\mathbf{,}\mathbf{200}\mathbf{-}\mathbf{4}{\mathit{Q}}_{\mathbf{1}}\mathbf{+}\mathbf{8}{\mathit{Q}}_{\mathbf{2}}\mathbf{-}\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{8}{\mathbf{Q}}_{\mathbf{1}}\mathbf{+}\mathbf{4}{\mathbf{Q}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{0} \\ \mathbf{-}\mathbf{12}{\mathit{Q}}_{\mathbf{1}}\mathbf{+}\mathbf{4}{\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{0} \end{gathered}$$

Solving for Q2:

${\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{3}{\mathit{Q}}_{\mathbf{1}}$

Having the value of Q2which only has the term Q1, it can be substituted in the function that maximizes the production of facility 1 to solve and obtain the value of Q1:

$$\begin{gathered} \mathbf{1}\mathbf{,}\mathbf{200}\mathbf{-}\mathbf{4}{\mathit{Q}}_{\mathbf{1}}\mathbf{+}\mathbf{8}\mathbf{\left(}\mathbf{3}{\mathbf{Q}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{20}{\mathit{Q}}_{\mathbf{1}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{Q}}_{\mathbf{1}}\mathbf{=}\mathbf{1200}\mathbf{/}\mathbf{20} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{60} \end{gathered}$$

Now, substituting the value of Q1on the function that maximize the quantity for facility 2 to obtain Q2:

$$\begin{gathered} \mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{8}\mathbf{\left(}\mathbf{60}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{Q}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{680}\mathbf{/}\mathbf{4} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{420} \end{gathered}$$

Profit maximizing price and benefits

The profit-maximizing price of both facilities can be determined by substituting the values of Q1and Q2on the inverse demand function. Thus,

$$\begin{gathered} \mathit{Q}\mathbf{=}\left(Q_1+Q_2\right) \\ \mathit{P}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{4}\mathit{Q} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{4}\left(Q_1+Q_2\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{200}\mathbf{+}\mathbf{4}\left(60+420\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2020} \end{gathered}$$

The benefit of the electric utility company can be calculated as the Total Revenue less Total Cost:

$$\begin{gathered} \mathit{B}\mathit{e}\mathit{n}\mathit{e}\mathit{f}\mathit{i}\mathit{t}\mathbf{=}\mathit{T}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathbf{}\mathit{R}\mathit{e}\mathit{v}\mathit{e}\mathit{n}\mathit{u}\mathit{e}\mathbf{-}\mathit{T}\mathit{o}\mathit{t}\mathit{a}\mathit{l}\mathbf{}\mathit{C}\mathit{o}\mathit{s}\mathit{t} \\ \mathit{B}\mathit{e}\mathit{n}\mathit{e}\mathit{f}\mathit{i}\mathit{t}\mathbf{=}\mathbf{(}\mathbf{P}\mathbf{\times }\mathbf{Q}\mathbf{)}\mathbf{-}\mathbf{\{}\mathbf{C}\left(Q_1\right)\mathbf{+}\mathbf{C}\left(Q_2\right)\mathbf{\}} \\ \mathit{B}\mathit{e}\mathit{n}\mathit{e}\mathit{f}\mathit{i}\mathit{t}\mathbf{=}\mathbf{\{}\mathbf{2020}\mathbf{\times }\left(60+420\right)\mathbf{\}}\mathbf{-}\mathbf{\{}\mathbf{8}\mathbf{,}\mathbf{000}\mathbf{+}\mathbf{6}\left(60\right)\mathbf{+}\mathbf{6}\mathbf{,}\mathbf{000}\mathbf{+}\mathbf{3}\left(420\right)\mathbf{\}} \\ \mathit{B}\mathit{e}\mathit{n}\mathit{e}\mathit{f}\mathit{i}\mathit{t}\mathbf{=}\mathbf{969}\mathbf{,}\mathbf{600}\mathbf{-}\mathbf{15}\mathbf{,}\mathbf{620} \\ \mathit{B}\mathit{e}\mathit{n}\mathit{e}\mathit{f}\mathit{i}\mathit{t}\mathbf{=}\mathbf{953}\mathbf{,}\mathbf{980} \end{gathered}$$

Thus, the profit-maximizing output at the facility 1 is 60 units and at the facility 2 is 420 units. The optimal price is $2020. The company’s profit is $969,980.