Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 20H / ii / Solution

Capital and labour are nonnegative inputs. On the budget line, $0\le K\le b$ and $L=(b-K)/w$. Both endpoints give zero output, whereas every interior point has positive output, so the maximum is interior. The Lagrangian with the sign convention of part (i) isIts stationarity equations are $\alpha K^{\alpha -1}{L}^{\beta }=\lambda$ and $\beta K^{\alpha }{L}^{\beta -1}=w\lambda$. Dividing by the positive output gives $\alpha /K=\lambda /f$ and $\beta /L=w\lambda /f$, hence $wL=(\beta /\alpha )K$. Together with the budget constraint this yieldsTo verify the unique global maximum, maximize the logarithm of the Cobb–Douglas production function along the budget line:Its unique stationary point is therefore the unique maximum, including comparison with the zero-output boundary points.

Writing $r=\alpha +\beta$ and $C=(\alpha /r{)}^{\alpha }(\beta /(wr){)}^{\beta }$, the optimal output is $\varphi (b)=C{b}^r$. The Lagrange multiplier obtained from the first stationarity equation is $\lambda =\alpha \varphi /K_{\ast }=r\varphi /b$. Direct differentiation verifies the envelope theorem identity:For clarity, the global dual-supremum premise of part (i) need not hold for every exponent pair allowed here. If $r>1$, along $(K,L)=(t{K}_{\ast },t{L}_{\ast })$ the unrestricted Lagrangian is $t^r\varphi (b)+\lambda b(1-t)$, which tends to infinity for every finite $\lambda$. Thus the verification above is by constrained stationarity and direct differentiation, and remains valid even when that stronger global Lagrangian duality representation fails.