Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 20H (Optimization)

For fixed $b$, put ${\mathcal{L}}_b(x)=f(x)+\lambda (b{)}^T(b-g(x))$. Feasibility gives ${\mathcal{L}}_b(\bar{x}(b))=f(\bar{x}(b))=\varphi (b)$. The assumed Lagrangian duality representation says this value is the unrestricted supremum of ${\mathcal{L}}_b$, so $\bar{x}(b)$ is also an unconstrained global maximizer of this continuously differentiable function. Its first derivative vanishes:Differentiate the identity $g(\bar{x}(b))=b$ using the assumed smooth dependence of the optimizer:The chain rule now gives the envelope theoremThusEach Lagrange multiplier is the marginal change in optimal value per unit relaxation of its constraint. The proof only uses the resulting stationarity and smooth feasibility identities; it does not require differentiating the maximizer explicitly or differentiating the multiplier.

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.