Example: Suppose that the short-run production function of certain cut-flower firm is given by:
𝑸=𝟒𝑲𝑳−𝟎.𝟔𝑲^𝟐-0.1𝑳^𝟐 :where Q is quantity of cut-flower produced, L is labour input and K is fixed capital input (K=5).
Determine the average product of labour (APL) function.
At what level of labour does the total output of cut-flower reach the maximum?
What will be the maximum achievable amount of cut-flower production?
To determine the average product of labor (APL) function, we need to divide the total product of labor (Q) by the amount of labor input (L).
APL = Q / L = (4KL - 0.6K^2 - 0.1L^2) / L = 4K - 0.6K^2/L - 0.1L
To find the level of labor (L) at which the total output of cut-flower reaches the maximum, we need to take the derivative of the production function with respect to labor input (L) and set it equal to 0.
dQ/dL = 0 gives us 4K - 0.2L = 0
Solving for L, we get L = 20K.
Substituting L = 20K back into the production function, we can find the maximum achievable amount of cut-flower production.
Q_max = 4KL - 0.6K^2 - 0.1L^2
= 4K(20K) - 0.6K^2 - 0.1(20K)^2
= 80K^2 - 0.6K^2 - 400K^2
= 79.4K^2
So, the maximum achievable amount of cut-flower production is 79.4K^2.
To determine the average product of labor (APL) function, we need to divide the total product (TP) by the amount of labor input (L). The total product is given by the production function, 𝑸 = 4𝑲𝐿 − 0.6𝑲^2 − 0.1𝐿^2.
Step 1: Find the total product (TP)
Substitute the fixed capital input (K = 5) into the production function:
𝑸 = 4(5)𝐿 − 0.6(5)^2 − 0.1𝐿^2
Simplify the equation:
𝑸 = 20𝐿 − 15 − 0.1𝐿^2
Step 2: Calculate the average product of labor (APL)
Divide the total product (TP) by the labor input (L):
𝐴𝑃𝐿 = 𝑄/𝐿
Plugging in the expression for TP:
𝐴𝑃𝐿 = (20𝐿 − 15 − 0.1𝐿^2)/𝐿
Simplify the equation:
𝐴𝑃𝐿 = 20 − (15/𝐿) − 0.1𝐿
The average product of labor (APL) function can be represented as 𝐴𝑃𝐿 = 20 − (15/𝐿) − 0.1𝐿.
To find the level of labor input that yields the maximum total output of cut-flowers, we need to differentiate the total product (TP) function with respect to labor (L) and set the derivative equal to zero.
Taking the derivative of TP = 20𝐿 − 15 − 0.1𝐿^2 with respect to L:
𝑑𝑇𝑃/𝑑𝐿 = 20 − 0.2𝐿 = 0
Solving for L:
20 − 0.2𝐿 = 0
0.2𝐿 = 20
𝐿 = 20/0.2
L = 100
Therefore, at a labor input of 100, the total output of cut-flowers is maximized.
To find the maximum achievable amount of cut-flower production, substitute the value of L = 100 into the production function:
𝑄 = 4(5)(100) − 0.6(5)^2 − 0.1(100)^2
Simplify the equation:
𝑄 = 2000 − 15 − 0.1(10000)
𝑄 = 1985
Therefore, the maximum achievable amount of cut-flower production is 1985 units.