Question 12.13: 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.)
The objective is to find the value of K which maximizes Q = 12K^{0.4} (160 − 8K)^{0.4} . In Example 12.9, first-order conditions were satisfied when
\frac{dQ}{dK} = \frac{768 – 76.8K}{(160 – 8K)^{0.6}K^{0.6}}
which holds when K = 10.
To derive d²Q/dK² let u = 768 − 76.8K and v = (160 − 8K)^{0.6} K^{0.6} . Therefore,
\frac{du}{dK} = −76.8 (1)
and, using the product rule,
\frac{dv}{dK} = (160 − 8K)^{0.6} 0.6K^{-0.4} + K^{0.6} 0.6(160 − 8K)^{-0.4} (−8)
= \frac{(160 – 8K)0.6 – 4.8K}{K^{0.4}(160 – 8K)^{0.4}}
= \frac{96 – 9.6K}{K^{0.4}(160 – 8K)^{0.4}} (2)
Therefore, using the quotient rule and substituting (1) and (2)
\frac{d^2Q}{dK^2} =
\frac{(160 − 8K)^{0.6}K^{0.6}(−76.8) − (768 − 76.8K) \frac{96 − 9.6K}{K^{0.4}(160 − 8K)^{0.4}}}{(160 − 8K)^{1.2}K^{1.2}}
= \frac{(160 − 8K)K(−76.8) − 76.8(10 – K)9.6(10 – K)}{(160 − 8K)^{1.6}K^{1.6}}
At the stationary point when K = 10 several terms become zero, giving
\frac{d^2Q}{dK^2} = \frac{-76.8(800)}{(800)^{1.6}} < 0
Therefore, the second-order condition for a maximum is satisfied when K = 10.