Answer to Question #234546 in Microeconomics for Jaysling

Question

1. The following production function has been estimated for wheat: W=4F^0.5 . L^0.25
Where, W=wheat yield (kg/ha), F=fertiliser (kg/ha) and L=Labour (hours/ha)

a) derive an expression for the least cost expansion path given that fertiliser cost R12 per kg and Labour cost R6 per hour.
b)if the wheat producer has a budget of R9000 to spend on fertiliser and Labour, how much of input should be used and what is the quantity of wheat he will produce?
c) Derive expression for the value of marginal product (VMP) of fertiliser and Labour given that wheat sells at R72 per kg.

Answer 1

Given information:

$W = 4 F^{0.5} L^{0.25}$

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

----

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

i.e,$PF = 12 and PL = 6$

part (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

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

$MPL = 0.5(4) F^{0.5}L^{0.25-1}$

$MPL = F^{0.5} L^{-0.75}$

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

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At cost minimization point following condition must holds true:

$(MPL / MPF) = (PL / PF)$

$[\frac{(\frac{F^{0.5}}{L^{0.75}})}{2(\frac{L^{0.25}}{F^{0.5}})} ] = (6 /12)$

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

$[\frac{F}{2L}] = 0.5$

$F=2L\times 0.5$

=>$F = L$   --------------> 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.

part (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

----

$W = 4 F^{0.5} L^{0.25}$

=$4(500)^{0.75}$

$=422.95 units$

which is optimuum quantity product

(part-c)

$now, VMP=P_{W} MPL$

now $MPL=1F^{1/2} L^{-3/4}$

or,$MPL=F^{1/2} L^{-3/4}$

$P_W=R72/kg$

$\therefore$ $VMP=P_WMPL$

=$(72F^{1/2})/(L^{3/4})$