The Second Largest Public Utility In The Nation Is The Sole Provider O
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 + 6Q1
2 and C2(Q2) = 6,000 + 3Q2
2 , respectively. Determine the profit-maximizing amounts of electricity to produce at the two facilities, the optimal price, and the utility company’s profits.
Paper For Above instruction
The utility company in southern Florida, serving as the sole electricity provider across 32 counties, faces the complex task of maximizing profits while satisfying the demand for electricity. The demand function, P = 1,200 – 4Q, illustrates that as total quantity (Q) increases, the price consumers are willing to pay decreases linearly. The company operates two generating facilities; Facility 1 produces Q1 kilowatts with a quadratic cost function C1(Q1) = 8,000 + 6Q1
2 , while Facility 2 produces Q2 kilowatts with C2(Q2) = 6,000 + 3Q2
2
. The combined production is Q = Q1 + Q2, and the goal is to determine the optimal production levels at both facilities, the corresponding market price, and the total profits for the utility.
To approach this problem, we begin with the revenue function, which depends on the total quantity supplied. Since the inverse demand function is P = 1,200 – 4Q, total revenue (TR) is given by TR = P × Q = (1,200 – 4Q) × Q = 1,200Q – 4Q
2
. This quadratic function reflects the decreasing price with increased total quantity.
The total cost (TC) is the sum of costs from both facilities: TC = C1(Q1) + C2(Q2) = (8,000 + 6Q1
2 ) + (6,000 + 3Q2 2 ) = 14,000 + 6Q1 2 + 3Q2 2
. The total profit (π) is thus:
π = Total Revenue – Total Cost = (1,200Q – 4Q
2
) – (14,000 + 6Q1
2 + 3Q2 2 ).
Since Q = Q1 + Q2, we express the profit function in terms of Q1 and Q2:
π = 1,200(Q1 + Q2) – 4(Q1 + Q2) 2 – 14,000 – 6Q1 2 – 3Q2
. Expanding, we obtain:
= 1,200Q1 + 1,200Q2 – 4(Q1
+ 2Q1Q2 + Q2
) – 14,000 – 6Q1
– 3Q2
. Simplifying further:
= 1,200Q1 + 1,200Q2 – 4Q1
– 8Q1Q2 – 4Q2
– 14,000 – 6Q1
– 3Q2
. Combine like terms:
= (1,200Q1 – 4Q1
2 – 6Q1
2
) + (1,200Q2 – 4Q2
2 – 3Q2
2
) – 8Q1Q2 – 14,000, which simplifies to:
π = 1,200Q1 – 10Q1
2 + 1,200Q2 – 7Q2
2 – 8Q1Q2 – 14,000.
To find the profit-maximizing production levels, we take the partial derivatives of π with respect to Q1 and Q2 and set them to zero:
∂π/∂Q1 = 1,200 – 20Q1 – 8Q2 = 0,
∂π/∂Q2 = 1,200 – 14Q2 – 8Q1 = 0.
This system of equations can be expressed as:
20Q1 + 8Q2 = 1,200,
8Q1 + 14Q2 = 1,200.
The matrix form is:
\[
\begin{bmatrix}
20 & 8 \\
8 & 14
\end{bmatrix}
\begin{bmatrix}
Q1 \\ Q2
\end{bmatrix} =
\begin{bmatrix}
1,200 \\ 1,200
\end{bmatrix}.
\]
Using methods such as substitution or matrix inversion, we solve for Q1 and Q2. Calculating the determinant:
Det = (20)(14) – (8)(8) = 280 – 64 = 216.
The inverse matrix is:
\[
\frac{1}{216}
\begin{bmatrix}
14 & -8 \\
\end{bmatrix}.
\]
Applying the inverse to the right-hand side: \[
Q1 = \frac{1}{216} \times (14 \times 1,200 - 8 \times 1,200) = \frac{1}{216} \times (16,800 - 9,600) = \frac{7,200}{216} = 33.33, \]
\[
Q2 = \frac{1}{216} \times (-8 \times 1,200 + 20 \times 1,200) = \frac{1}{216} \times (-9,600 + 24,000) = \frac{14,400}{216} = 66.67.
\]
The profit-maximizing production levels are approximately Q1 = 33.33 kilowatts and Q2 = 66.67 kilowatts. The total quantity supplied is Q = Q1 + Q2 ≈ 100.
The optimal price is determined by substituting Q into the inverse demand function:
P = 1,200 – 4Q = 1,200 – 4 × 100 = 1,200 – 400 = 800.
Calculating the total profit at these levels:
TR = P × Q = 800 × 100 = 80,000.
Total costs:
C1 = 8,000 + 6(33.33)
2
≈ 8,000 + 6 × 1,111.11 ≈ 8,000 + 6,666.67 = 14,666.67, C2 = 6,000 + 3(66.67)
2
≈ 6,000 + 3 × 4,444.44 ≈ 6,000 + 13,333.33 = 19,333.33.
Total costs = 14,666.67 + 19,333.33 = 34,000.
Profit = TR – Total Costs ≈ 80,000 – 34,000 = 46,000.
In conclusion, the utility should produce approximately 33.33 kilowatts at Facility 1 and 66.67 kilowatts at Facility 2. The resulting market price should be around $800 per kilowatt, resulting in an approximate profit of $46,000. These decisions optimize profit by balancing production costs with the demand-driven price, illustrating the strategic considerations monopolistic utilities must navigate (Varian, 2010; Pindyck & Rubinfeld, 2018).
References
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