a) To maximize the output we have to solve the equation:
"Q'=(6L\u00b2-0.4L\u00b3)'= 0"
"12L-1.2L\u00b2=0"
"L(12-1.2L)=0"
"L1=0"
"L2=10"
b)"MP=12L-1.2L\u00b2." To maximize we have to calculate: MP'=0, so
"12-2.4L=0"
"L=5."
c)"AP=6L-0.4L\u00b2" To maximize we have to calculate AP'=0, so
"6-0.8L=0"
"L=7.5"
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Thanks Teacher
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