a) To maximize the output we have to solve the equation:
Q′=(6L2−0.4L3)′=0
12L−1.2L2=0
L(12−1.2L)=0
L1=0
L2=10
b)MP=12L−1.2L2. To maximize we have to calculate: MP'=0, so
12−2.4L=0
L=5.
c)AP=6L−0.4L2 To maximize we have to calculate AP'=0, so
6−0.8L=0
L=7.5
Comments
Thanks Teacher
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