(a) Q = 0.4k^0.6 L^0.7 MP_k = d theta/d k = 04 times 0.6 k^0.6 - 1 L^0.7 = 0.24
ID: 3868720 • Letter: #
Question
(a) Q = 0.4k^0.6 L^0.7 MP_k = d theta/d k = 04 times 0.6 k^0.6 - 1 L^0.7 = 0.24 k^-0.4 L^0.7 the value of k = 200 and L = 2000 MP_k = 0.24 (300)^-0.4 (2000)^0.7 = 0.24 times 0.102 times 204.513 = 5.01 (b) MP_L = d theta/d L = 0.4 times 0.7 L^0.7 - 1 k^0.6 = 0.28 L^- 0.3 k^0.6 substituting the value of k = 300 & L = 2000 MP_L = 0.28(2000)^-0.3 (300)^0.6 = 0.28 times 0.102 times 30.639 = 0.88 (c) The estimated equation represents increasing returns to scale because 0.6 + 0.7 = 1.3 > 1 (d) Originally, when k = 300 and L = 2000 Q = 0.4 k^0.6 L^0.7 = 0.4 (300)^0.6 (2000)^0.7 = 0.4 times 30.639 times 204.513 =2506.43 If use of labor will increase by it means now labor quantity 2000 + (0.08 times 2000) = 2160 Q = 0.4(300)^0.6 (2160)^0.7 = 0.4 times 30.639 times 215.833 = 2645.16 Thus output will increase by 2645.16 - 2506 43 = 138.73Explanation / Answer
a)Q=0.4*(k^0.6)*(l^0.7)
MPk=do/dk=0.4*0.6*(k^0.6-1)*(L^0.7)
=0.24*(k^-0.4)*(L^0.7)
Substituting the value of K=300 and L=2000
MPk=0.24*(300^-0.4)*(2000^0.7)
=0.24*0.102*204.513
=5.01
b)MPl=do/dl=0.4*0.7*(L^0.7-1)*(k^0.6)
=0.28*(L^-0.3)*(k^0.6)
Substituting the value of k=300 and L=2000
MPl=0.28*(2000^-0.3)*(300^0.6)
=0.28*0.102*30.639
=0.88
c)The estimated equation represents increasing returns to scale because
0.6+0.7=1.3>1
d)Originally,when k=300 and L=2000
Q=0.4*(k^0.5)*(L^0.7)
=0.4*(300^0.6)*(2000^0.7)
=0.4*30.639*204.513
=2506.43
If use of labour will increase by 8%,it means
now labour quantity is
2000+(0.08*2000)=2160
Q=0.4*(300^0.6)*(2160^0.7)
=0.4*30.639*215.833
=2645.16
Thus output will increase bt
2645.16-2506.43=138.73 units.
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