Question #204454

5. Parexel Co. produces and sells cell phone batteries. Its production function is 𝒒 = 𝟑𝟐𝟏𝑲𝟎.𝟑𝑳𝟎.𝟑, while its cost function is 𝒄 = 𝟐𝟏𝟎𝑲 + 𝟗𝟎𝑳. Parexel can sell its batteries at 𝒑 = 𝟔𝟑.

a. Does this production function exhibit increasing, decreasing or constant returns to scale? Explain briefly.

[2]

b. Does this production function exhibit decreasing returns to capital? Explain briefly. [2]

c. Use implicit differentiation to find the marginal rate of technical substitution, 𝑴𝑹𝑻𝑺. [2]

d. Write down an expression for the isoquant if 𝒒 = 𝟐𝟎𝟒𝟖. [2]

e. Write down an expression for the labour elasticity of production, 𝒆𝑳. [2]

f. Which levels of 𝑳 and 𝑲 satisfy the the first-order and second-order conditions for the maximisation of

Parexel’s profit? [8]

g. Write down the levels of profit, revenue, cost and output at the profit maximising levels of 𝑲 and 𝑳. [2]


1
Expert's answer
2021-06-24T12:36:10-0400

a

q0=321K0.3L0.3q_0=321K^{0.3}L^{0.3}

multiply each input by m

q=321(m)0.6K0.3L0.3m0.6q0q=321(m)^{0.6}K^{0.3}L^{0.3}\\m^{0.6}q_0

i.e output increases by less than m, hence decreasing return to scale


b.


MPk=(321)(0.3)L0.3K0.7δMPkδk=(321)(0.3)(0.7)L0.3K1.7<0MP_k=(321)(0.3)L^{0.3}K^{-0.7}\\\frac{\delta MP_k}{\delta k}=(321)(0.3)(-0.7)L^{0.3}K^{-1.7}<0


This implies MPk is decreasing

hence Diminishing marginal return


c.

MRTS=δqδLδqδKMRTS=\frac{\frac{\delta q}{\delta L}}{\frac{\delta q}{\delta K}}


=(321)(0.3)K0.3L0.7(321)(0.3)K0.7L0.3=\frac{(321)(0.3)K^{0.3}L^{-0.7}}{(321)(0.3)K^{-0.7}L^{0.3}}


MRTS=KLMRTS=\frac{K}{L}


d.

q=2048    2048=321K0.3L0.3    K0.3L0.3    K0.3L0.3=6.4q=2048\\\implies2048=321K^{0.3}L^{0.3}\\\implies K^{0.3}L^{0.3}\\\implies K^{0.3}L^{0.3}=6.4 Isoquant


e.

labor elasticity

L=(lq)(dqdl)\in_L=(\frac{l}{q})(\frac{dq}{dl})

=l(321)K0.3L0.3((321)(+0.3)K0.3L0.7)=\frac{l}{(321)K^{0.3}L^{0.3}}((321)(+0.3)K^{0.3}L^{-0.7})


=0.3\in=0.3


f.

Profit will be mximized when

MPk=210

(321)(0.3)K0.7L0.3=210...................................(1)(321)(0.3)K^{-0.7}L^{0.3}=210...................................(1)

amd MPl=90

(321)(0.3)K0.3L0.7=90..................................(2)(321)(0.3)K^{0.3}L^{-0.7}=-90..................................(2)

by (1)(2)\frac{(1)}{(2)}

LK=219=73\frac{L}{K}=\frac{21}{9}=\frac{7}{3}

L=73KL=\frac{7}{3}K

from cost function

2048=321(K0.3)(73)0.3K0.3    K0.6=8.222048=321(K^{0.3})(\frac{7}{3})^{0.3}K^{0.3}\\\implies K^{0.6}=8.2\sqrt2


K=33.52L=78.22K=33.52\\L=78.22


g.

cost(C)=210K+90L=210(33.52)+90(78.23)=1079cost (C)=210K+90L\\=210(33.52)+90(78.23)= 1079

Revenue=P×Q=(63)(2048)=129024profit=TRTC    1290241079=14945Revenue=P\times Q= (63)(2048) =129024\\profit=TR-TC\implies129024-1079=14945

From the question q is given as 2048 therefore,

output=2048\\output=2048


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