Answer to Question #233075 in Microeconomics for Anna Paula Quimbra

Question

The following production function has been estimated for wheat:

W = 4 F^(0.5). L^(0.25)

where, W = Wheat yield (kg/ha), F = Fertiliser (kg/ha) and L = Labour (hours/ha)

A) If the wheat producer has a budget of R9 000 to spend on fertiliser and labour, how much of each input should he use and what is the quantity of wheat he will produce? [4]

B) Derive expressions for the value of marginal product (VMP) of fertiliser and labour given that wheat sells at R72 per kg.

Answer 1

complete question

The following production function has been estimated for wheat:

W = 4 F^0.5 L^0.25

where, W = Wheat yield (kg/ha), F = Fertiliser (kg/ha) and L = Labour (hours/ha)

a) Derive the expression for the least-cost expansion path given that fertilizers cost R12 per kg and labour cost R6 per hour

b) If the wheat producer has a budget of R9 000 to spend on fertiliser and labour, how much of each input should he use and what is the quantity of wheat he will produce?

c) Derive expressions for the value of marginal product (VMP) of fertiliser and labour given that wheat sells at R72 per kg.

Solution

A.

W = 4 F^0.5 L^0.25

Differentiate W w.r.t F to get the marginal product of fertilizer

$=> MPF = \frac{dW}{ dF}$

=> MPF = 0.5(4) F^{0.5 -1} L^0.25

=> MPF = 2 F^-0.5 L^0.25

$=> MPF = 2(\frac{L^{0.25} }{ F^{0.5}})$

and

Differentiate W w.r.t to get the marginal product of labor

$=> MPL = \frac{dW}{dL}$

=> MPL = 0.25(4)F^0.5 L^{0.25 -1}

=> MPL = F^0.5 L^-0.75

$=> MPL = \frac{F^{0.5} }{ L^{0.75}}$

At cost minimization point following condition must holds true:

$(\frac{MPL }{ MPF}) = (\frac{PL}{ PF})$

$=> [\frac{(\frac{F^{0.5} }{ L^{0.75}}) }{ 2(\frac{L0.25 }{ F0.5})} ] = (\frac{6}{12})$

$=> [\frac{(F^{0.5} \times F^{0.5}) }{ 2(L^{0.25} \times L^{0.75})}] = 0.5$

$=> (\frac{F }{ 2L}) = 0.5$

$=> F = 2L \times0.5$

=> F = L   --------------> The expression of least cost expansion path

note: Expansion path is the locus of all combination of inputs (i.e., Fertilizer and labor) that minimize the cost of production.

 

B.

The fertilizers cost R12 per kg and labour cost R6 per hour

i.e, PF = 12 and PL = 6

The budget to spend on fertilizer and labour is 9000

Cost constraint:

6L + 12F = 9000

Substitute F = L (i.e., equation path)

=> 6L + 12L = 9000

=> 18L = 9000

$=> L = (\frac{9000 }{ 18})$

=> L = 500

and 

F = L 

=> F = 500

500 kg of fertilizer and 500 labor hour will be used by wheat farmer

W = 4 F^0.5 L^0.25

Put F= 500 and L=500

=> W = 4 (500)^0.5 (500)^0.25

^=> W = 422.9

=> W = 423

423 Kg of wheat will be produced

C.

The marginal product of labor and marginal product of fertilizer are given below. It is calculated in part (a)

$MPL = \frac{F^{0.5} }{ L^{0.75}}$

$MPF = 2(\frac{L0.25 }{ F0.5})$

 of marginal product = Marginal product * Price of output

The price of wheat is 72 per kg 

The value of marginal product of labor

=> VMPL = MPL * Price of wheat

$=> VMPL = (\frac{72F^{0.5} }{ L^{0.75}})$

The value of marginal product of fertilizer

=> VMPF = MPF * Price of wheat

$=> VMPF = 2(\frac{L^{0.25}}{ F^{0.5}}) \times 72$

$=> VMPF = (\frac{144L^{0.25}} { F^{0.5}})$