a)
Given
"Q =L^{0.75} K^{0.5} ......... (1)"
Price of labor = 5 birr per day
Price of capital = 10 birr per day
Total cost budget = 400 birr per day
Budget line or isocost line
"5L+10K=400 ......(2)"
Maximize "Q =L^{0.75} K^{0.5}"
Subject to constraints "5L+10K=400"
Using lagrangian method
"X =L^{0.75} K^{0.5} +\u03bb(400-5L-10K)"
FOC
"\\frac{dX}{dL}=0, \\frac{dX}{dK}=0, \\frac{dX}{d\u03bb}=0"
Differentiating Lagrangian function with respect to L
"\\frac{dX}{dL}=0\\\\\u21d20.75L^{\u22120.25}K^{0.5}=5\u03bb ..........(3)\\\\\nNow\\\\\n\n\\frac{dX}{dK}=0\\\\\u21d20.5L^{0.75}K^{\u22120.5}=10\u03bb ..........(4)\\\\\nand \\\\\n\n\\frac{dX}{d\u03bb}=0\\\\\u21d25L+10K=400 ..........(2)"
Dividing eq 3 and 4
"\u21d2\\frac{0.75L^{\u22120.25}K^{0.5}}{0.5L^{0.75}K^{\u22120.5}}=\\frac{5\u03bb}{10\u03bb}\\\\\u21d2\\frac{3 K}{2L}=\\frac{1}{2}\\\\\u21d2K=\\frac{L}{3}"
Putting value of K in eq 2
"5L+10(\\frac{L}{3}) = 400\\\\\u21d2\\frac{15+10}{3}L=400\\\\\u21d2L=400\u00d7\\frac{3}{25}\\\\\u21d2L=48\\\\ \nAnd\\space K = \\frac{48}{3} = 16"
The optimal level of L is 48 and K is 16 that maximizes the output.
(b)
Maximum output will be
"Q^*=48^{0.75}\u00d716^{0.5}\\\\\u21d2Q^*= 72.944"
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