Question 12.9: At what level of K is the function Q = 12K^0.4(160 − 8K)^0.4......
At what level of K is the function Q = 12K^{0.4}(160 − 8K)^{0.4} at a maximum? (This is Example 11.1 (reworked) which was not completed in the last chapter.)
We need to differentiate the function Q = 12K^{0.4}(160 − 8K)^{0.4} to check the first-order condition for a maximum. To use the product rule, let
u = 12K^{0.4} and v = (160 − 8K)^{0.4}
and so
\frac{du}{dK} = 4.8K^{-0.6} and \frac{dv}{dK} = 0.4(160 − 8K)^{-0.6}(−8)
= −3.2(160 − 8K)^{-0.6}
Therefore,
\frac{dQ}{dK} = 12K^{0.4}(−3.2)(160 − 8K)^{-0.6} + (160 − 8K)^{0.4} 4.8K^{-0.6}
= \frac{−38.4K + (160 − 8K)4.8}{(160 − 8K)^{0.6}K^{0.6} }
= \frac{768 – 76.8K }{(160 − 8K)^{0.6}K^{0.6} } (1)
Setting (1) equal to zero for a stationary point must mean
768 − 76.8K = 0
K = 10