Question

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

Derive an expression for the least-cost expansion path given that fertiliser costs R12 per kg and labour costs R6 per hour. [4]

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]

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

Answer 1

$\frac{∂W/∂L}{∂W/∂F}=\frac{P_L}{P_F} \\ or \; \frac{MP_L}{MP_F}= \frac{P_L}{P_F} \\ \frac{F^{1/2}L^{-3/4}}{2F^{-1/2}L^{1/4}} = \frac{6}{12} \\ \frac{L^*}{F^*} = 1 \\ or \; L^*=F^*$

which is the expression for the least cost expansion path

Now,

$C=P_fF + P_lL \\ C=12F+6L \\ C=9000 \\ L^*=F^* \\ 18L^*=9000 \\ L^* = 500 \;units \\ F^*=500 \;units$

The total production

$W^* = 4F^{*1/2}L^{*3/4} \\ =4(500)^{3/4} \\ = 422.95 \;units$

which is the optimum quantity produced

Now, $VMP =P_W \times MP_L$

$MP_L= F^{1/2}L^{-3/4} \\ P_W = R72 \;per \;kg \\ VMP = P_W \times MP_L \\ = \frac{72F^{1/2}}{L^{3/4}}$