Consider the following production function.
TPL = 12L2 – 0.8L3
Determine the marginal product function(MPL)
Determine the average product function (APL)
Find the value of L that maximizes TPL
Find the value of L that maximizes APL
Find the value of L that maximizes MPL
"TP_L=12L^2 -0.8L^3 \\\\\n\ni) \\; MP_L= \\frac{\u2202(TP_L)}{\u2202(L)} \\\\\n\nMP_L= 2 \\times 12L^{(2-1)} \u2013 3 \\times 0.8L^{3-1} \\\\\n\nMP_L = 24 L -2.4L^2 \\\\\n\nii) \\; AP_L= \\frac{TP_L}{L} \\\\\n\nAP_L= \\frac{12L^2 -0.8L^3}{L} \\\\\n\nAP_L= 12L -0.8L^2"
iii) Maximozation of "TP_L"
Put "\\frac{\u2202(TP_L)}{\u2202(L)} = 0" for maximum value
"24L-2.4L^2=0 \\\\\n\nL(24-2.4L)=0 \\\\\n\n24=2.4L \\\\\n\nL= \\frac{24}{2.4}=10 \\;units"
iv) Put "\\frac{\u2202(AP_L)}{\u2202(L)} =0" , for maximum value
"AP_L= 12L -0.8L^2 \\\\\n\n\\frac{\u2202(AP_L)}{\u2202(L)} = 12 -1.6L \\\\\n\n12-1.6L=0 \\\\\n\nL=\\frac{12}{1.6}= 7.5 \\;units"
v) Put "\\frac{\u2202(MP_L)}{\u2202(L)}=0" for maximum value
"MP_L=24L -2.4L^2 \\\\\n\n\\frac{\u2202(MP_L)}{\u2202(L)} = 24-4.8L \\\\\n\n24-4.8L=0 \\\\\n\nL=\\frac{24}{4.8}= 5 \\;units"
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