Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 31 / 2 / Solution

Choose the sign conventionThis is the optimization Lagrangian; equality-constraint Lagrange multipliers are unrestricted in sign. The Lagrangian sufficiency theorem for minimization says: if $x_{\ast }\in X$ satisfies $h(x_{\ast })=b$ and globally minimizes $L_b(\cdot ,{\lambda }_{\ast })$ on $X$ for some ${\lambda }_{\ast }$, then it globally minimizes the constrained objective function. Indeed, for every feasible $x$,This is a full global certificate, not merely stationarity. No convexity is needed for the implication, although convexity commonly supplies the required global optimization Lagrangian minimum.

Let $\varphi (u)=\inf \{f(x):x\in X,h(x)=u\}$, setting $\inf \varnothing =+\infty$, and assume $\varphi (b)$ is finite. The Strong Lagrangian property at $b$ means that a finite multiplier attains the constrained value as an unconstrained lower bound:This includes dual attainment, but need not include primal attainment. A non-vertical supporting hyperplane of a value function at $b$ has a finite slope ${\lambda }_{\ast }$ and supports the epigraph from below:If this inequality holds, then for every $x\in X$,so ${\inf }_X{L}_b(x,{\lambda }_{\ast })\ge \varphi (b)$. Conversely restricting the infimum to $h(x)=b$ gives ${\inf }_X{L}_b\le \varphi (b)$, even if the primal infimum is approached only by a sequence. Thus equality holds. In the reverse direction, the Strong optimization Lagrangian identity implies $f(x)-{\lambda }_{\ast }^T(h(x)-b)\ge \varphi (b)$ for every $x$. Taking the infimum over $h(x)=u$ yields the supporting inequality, with empty fibres causing no problem. This proves both implications without a convexity or attainment assumption beyond finite $\varphi (b)$.

For the advertising calculation, measure the budget in thousands of pounds by $B=a/1000$. Requiring the full budget to be spent gives $3x+y=B$, with $x,y\ge 0$. Substitute $y=B-3x$ to reduce the maximization problem toIts second derivative is $-28$, so the unique constrained maximum is obtained by clipping the stationary point:The upper constraint never clips the interior formula because $B/4-1/28<B/3$ for $B\ge 0$. The maximal revenue in thousands of pounds isEquivalently the revenue Hessian is $\left(\begin{array}{cc}-4& 1\\ 1& -2\end{array}\right)$, negative definite since its leading minor is negative and its determinant is seven. Thus the full-budget affine feasible set has a unique global maximum.

At the two requested budgets,The units in the middle columns are minutes; the last column is thousands of pounds. Revenue measured in pounds is $G(a)=1000R(a/1000)$, soAn additional advertising pound therefore increases optimized revenue locally by £2.50 at the smaller budget and £0.25 at the larger budget. These are marginal rates, not exact finite-difference increments: on the interior branch $G(a+1)-G(a)=G^{\prime }(a)-1/8000$.

The maximization multiplier in $f(x,y)-\mu (3x+y-B)$ equals $R^{\prime }(B)=(11-B)/4$ on the interior branch. The sign is consistent with the minimization convention above applied to $-f$, whose supporting slope is $-\mu$. If spending is only capped by the budget, rather than required to equal it, these same answers apply at the two budgets. Beyond £11,000 the unconstrained maximum $(x,y)=(19/7,20/7)$ is affordable, so further allowed spending is unused and the capped-budget value is constant; forced full spending would instead lower revenue.