a) Based on the current inputs used, compute the level of output
The production function is:
"Q = 30L^{0\u22c585}K^{0.20}\\\\[0.3cm]\nQ = 30(25)^{0.85}(20)^{0.20}\\\\[0.3cm]\nQ = 842.5123"
b) Compute the current total costs
The total cost equation is:
"C = 75K + 50L\\\\[0.3cm]\nC = 75(20) + 50(25)\nC = 2,750"
c) Given the current input usage, is the first operation efficiently?
At the optimal production:
"\\dfrac{MP_L}{MP_K} = \\dfrac{w}{r}\\\\[0.3cm]\nMP_L = \\dfrac{\\delta Q}{\\delta L} = 25.5L^{-0.15}K^{0.20}\\\\[0.3cm]\n\\rightarrow MP_L = 25.5(25)^{-0.15}(20)^{0.20} = 26.6454\\\\[0.3cm]\nMP_K = \\dfrac{\\delta Q}{\\delta K} = 6L^{0.85}K^{-0.80}\\\\[0.3cm]\n\\rightarrow MP_K = 6(25)^{0.85}(20)^{-0.80} = 8.4251"
Therefore:
"\\dfrac{MP_L}{MP_K} = \\dfrac{26.6454}{8.4251} = 3.1626\\\\[0.3cm]\n\\dfrac{w}{r} = \\dfrac{50}{75} = 0.6667"
Since "\\dfrac{MP_L}{MP_K}>\\dfrac{w}{r}" , the firm is not producing the efficient level of output.
d) Derive the equation path equation
The equation path or the expansion path is derived from the cost minimization problem.
"\\dfrac{MP_L}{MP_K} = \\dfrac{w}{r}\\\\[0.3cm]\n\\dfrac{25.5L^{-0.15}K^{0.20}}{6L^{0.85}K^{-0.80}} = \\dfrac{50}{75}\\\\[0.3cm]\n\\color{red}{\\dfrac{K}{L} = \\dfrac{8}{51}}"
e) Does the production function exhibits increasing, constant or decreasing returns to scale? Explain.
For a Cobb-Douglas production function "Q = aL^aK^b" , the firm experiences increasing returns to scale if "a + b = 1" , decreasing returns to scale if "a + b<0" and constant returns to scale if "a + b = 1" .
Our production function is:
"Q = 30L^{0\u22c585}K^{0.20}\\\\[0.3cm]\n\\rightarrow a = 0.85,\\; b = 0.20"
Therefore:
"\\rightarrow a + b = 0.85 + 0.20 = 1.05>1"
The production function exhibits increasing returns to scale.
f) Determine the percentage increase in output if both labour and capital are each increased by 15%
"Q = 30L^{0\u22c585}K^{0.20}\\\\[0.3cm]\nQ_1 = 30(1.15L)^{0.85}(1.15K)^{0.20}\\\\[0.3cm]\nQ_1 = 1.1581(30L^{0\u22c585}K^{0.20})\\\\[0.3cm]\nQ_1 = 1.1581Q"
This means that the quantity has increased by a percentage of:
"\\Delta Q = 1.1581 - 1\\\\[0.3cm]\n\\Delta Q = 0..1581 = 15.81\\%"
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