Question #209418

i. Suppose that the production function for compact disc player is

𝑄=100𝐿0.6𝐾0.4

Where 𝑄 is the total output, 𝐿 is the quantity of labor employed, and 𝐾 is the quantity of capital in place.

a) Calculate TP, AP, and MP for the sixth, seventh and eighth units of labour employed if capital is fixed at 240 units.


Expert's answer

Given 

Q=100L0.6K0.4.......(1)Q= 100L^{0.6}K^{0.4} ....... (1)

Where Q is total output, L is the quantity of labor employed. K is quantity of capital. 

If capital is fixed at K=240K=240

So 

Q=100L0.62400.4Q= 100L^{0.6} 240^{0.4}

Q=895.54L0.6......(2)Q= 895.54 L^{0.6} ...... (2)


Average product of labor, 

Dividing output function by L

AP(L)=895.54L(0.4).......(3)AP(L) = 895.54 L^{(-0.4)} ....... (3)

Marginal product of labor, differentiate equation 2 w r t L

MP(L)=TP(n)TP(n1)..........(4)MP(L) =TP(n)-TP(n-1) .......... (4)

 

When L=6L = 6

Q=895.54(60.6)Q=2624.07Q= 895.54{ (6^{0.6})}\\ Q= 2624.07

So total product is2624.072624.07

Average product 

AP=895.54(6)(0.4)=437.345AP= 895.54 (6)^{(-0.4)}=437.345

Marginal product of laborMP(6)=Q(6)Q(5)=2624.07895.54(50.6)=271.91MP(6) =Q(6)-Q(5) =2624.07-895.54 (5^{0.6}) =271.91


When labor is L= 7 

Total product 

Q=895.54(70.6)Q=2878.35Q= 895.54 (7^{0.6})\\ Q= 2878.35

So total product is 2878.352878.35

Average product 

AP=895.54(7)(0.4)=411.19AP= 895.54 (7)^{(-0.4)}=411.19

Marginal product of labor

MP(7)=TP(7)TP(6)=254.28MP(7) =TP(7)-TP(6) =254.28

When L = 8

Total product 

Q=895.54(80.6)Q=3118.45Q= 895.54 (8^{0.6}) \\ Q= 3118.45

So total product is 3118.453118.45

Average product 

AP=895.54(8)(0.4)=389.81AP= 895.54 (8)^{(-0.4)}=389.81

Marginal product of labor

MP(8)=TP(8)TP(7)=240.1MP(8) =TP(8)-TP(7)= 240.1


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