Given the production function y = min (z₁- 2, z₂- 4), the efficient number of units of output when there are 20 units of z₁and 30 units of z₂available is A)30 B)18 C)26 D)13

Question

Quizplus · Accepted Answer

To find the efficient number of units of output, we need to compare the output produced by each factor and choose the minimum one. 
For the first factor: zS1U1B11S1U1B0- 2, we have 20 units of zS1U1B11S1U1B0 available, so the output produced by this factor is:
20*(S1U1B11)^1*(S1U1B0)^1 = 20*S1U1B11*S1U1B0
For the second factor: zS1U1B12S1U1B0S1U1B1 S1U1B0- 4, we have 30 units of zS1U1B12S1U1B0 available, so the output produced by this factor is:
30*(S1U1B12)^1*(S1U1B0)^1*(S1U1B1)^1*(S1U1B0)^(-4) = 30*S1U1B12*S1U1B1/S1U1B0^4
To maximize the output, we have to choose the minimum output produced by these two factors, which is the first one:
20*S1U1B11*S1U1B0
To find the efficient number of units of output, we need to set this expression equal to y (the total output) and solve for S1U1B11 and S1U1B0:
y = min (20*S1U1B11*S1U1B0, 30*S1U1B12*S1U1B1/S1U1B0^4)
20*S1U1B11*S1U1B0 = y
S1U1B0 = (y/20)^(1/2)
30*S1U1B12*S1U1B1/S1U1B0^4 = y
S1U1B1 = (y*S1U1B0^4)/(30*S1U1B12)
Substituting S1U1B1 and S1U1B0 back into the first equation:
S1U1B11 = y/(20*S1U1B0)
S1U1B11 = y/[20*(y/20)^(1/2)]
S1U1B11 = (y/20)^(1/2)
Therefore, the efficient number of units of output is y = 20*S1U1B11*S1U1B0 = 20*(y/20)^(1/2)*(y/20)^(1/2) = 20*(y/20) = y = 20*S1U1B11*S1U1B0 = 20*(y/20)^(1/2)*(y/20)^(1/2) = 20*(y/20) = y = 360, which gives us an efficient number of units of output of 18 (y = 20*S1U1B11*S1U1B0 = 20*(2/3)*(3/2)). Therefore, the best choice is B.

Given the Production Function Y = Min (Z1- 2, Z2

Given the Production Function Y = Min (Z₁- 2, Z₂