Question 11.5: A firm operates with the production function Q = 4K^0.6L^0.4......
A firm operates with the production function Q = 4K^{0.6}L^{0.4} and buys inputs K and L at prices per unit of £40 and £15 respectively. What is the cheapest way of producing 600 units of output?
The output constraint is
600 = 4K^{0.6}L^{0.4}
Therefore
\frac{150}{K^{0.6}} = L^{0.4}
\left(\frac{150}{K^{0.6}} \right)^{2.5} = L
\frac{275,567.6}{K^{1.5}} = L (1)
The total cost of inputs, which is to be minimized, is
TC = 40K + 15L (2)
Substituting (1) into (2) gives
TC = 40K + 15(275,567.6)K^{-1.5}
Differentiating and setting equal to zero to find a stationary point
\frac{dTC}{dK} = 40 − 22.5(275,567.6){K^{-2.5}} = 0 (3)
40 = \frac{22.5(275,567.6)}{K^{2.5}}
{K^{2.5}} = \frac{22.5(275,567.6)}{40} = 155,006.78
K = 119.16268
Substituting this value into (1) gives
L = \frac{275,567.6}{(119.1628)^{1.5}} = 211.84478
This time we can check the second-order condition for minimization. Differentiating (3) again gives
\frac{d^2TC}{dK^2} = (2.5)22.5(275,567.6)K^{-3.5} > 0 for any K > 0
This confirms that these values minimize TC. We can also check that these values give 600 when substituted back into the production function.
Q = 4K^{0.6}L^{0.4} = 4(119.16268)^{0.6}(211.84478)^{0.4} = 600
Thus cost minimization is achieved when K = 119.16 and L = 212.84 (to 2 dp) and so total production costs will be
TC = 40(119.16) + 15(211.84) = £7,944