200x+100y=30000=>y=300−2x,x,y≥0 Then
F=F(x)=50x3/5(300−2x)2/5,0≤x≤150 Find the first derivative with respect to x
F′(x)=(50x3/5(300−2x)2/5)′
=50(53x−2/5(300−2x)2/5−52(2)x3/5(300−2x)−3/5)
=10x−2/5(300−2x)−3/5(3(300−2x)−4x)
=100x−2/5(300−2x)−3/5(90−x)
F′(x)=0=>100x−2/5(300−2x)−3/5(90−x)=0
=>x=90 Critical numbers: 0,90,150
F(0)=0
F(150)=0
F(90)=50(90)3/5(300−90)2/5=
=1500(1323)1/5≈6315.43
The maximum production level is $6315.43 and occurs when x=90 units of labor and y=120 units of capital.
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