Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 3 / 20C / Solution

For the Lagrangian sufficiency theorem, write a maximization problem as ${\max }_{x\in D}f(x)$ subject to $g_i(x)\le c_i$. Here $D$ may already incorporate simple constraints such as nonnegativity. Define the Lagrangian in optimizationThe theorem states that a feasible point $x_{\ast }$ is a global maximizer if there are such multipliers for which $x_{\ast }$ globally maximizes $L(\cdot ,\lambda )$ on $D$ and complementary slackness holds: ${\lambda }_i(g_i(x_{\ast })-c_i)=0$ for every $i$. To prove it, for any feasible $x$ observe thatThe first inequality uses feasibility and nonnegative multipliers, the second uses global maximization of the Lagrangian in optimization, and the last uses complementary slackness. Thus $x_{\ast }$ is globally optimal. No convexity hypothesis is needed for this implication; concavity is useful when establishing the required global maximization of $L$ from its gradient.

For the particular problem, put $d=e^{c_2}-1\ge 0$ and $R=c_1-2d\ge 0$. The logarithmic constraint is equivalent to $x_1\ge d$. The objective increases strictly with either variable, so at a maximum the resource constraint is tight: otherwise $x_1$ could be increased. HenceOn this interval the objective becomes $F(x_2)=(c_1-3x_2)/2+3\log (1+x_2)$, withIt increases up to $x_2=1$ and decreases thereafter, so the unique feasible maximizer isThe maximum value isThe two expressions agree at $R=3$. This includes the degenerate feasible set $R=0$, where $x_1^{\ast }=d$ and $x_2^{\ast }=0$.

To certify the result using the Lagrangian sufficiency theorem, take $D=[0,\infty {)}^2$ andwhere $\lambda ,\mu \ge 0$. SetSince $x_2^{\ast }\le 1$, $\lambda \ge 1/2$ and $\mu \ge 0$. These choices make both components of ${\nabla }_{x}L$ vanish at $x_{\ast }$: $1-2\lambda +\mu /(1+x_1^{\ast })=0$ and $3/(1+x_2^{\ast })-3\lambda =0$. The Lagrangian in optimization is concave on $D$, so the zero gradient proves global maximization there. The resource constraint is tight. When $R<3$ the logarithmic constraint is tight; when $R\ge 3$, $\mu =0$. Thus complementary slackness holds in every case, including both endpoints, and the theorem establishes the claimed optimum.