= Solution
Choose the sign convention
$$
L_b(x,\lambda)=f(x)-\lambda^T(h(x)-b),\qquad x\in X,\quad\lambda\in\mathbb R^m.
$$
This is the <optimization Lagrangian>; equality-constraint <Lagrange multipliers> are unrestricted in sign. The <Lagrangian sufficiency theorem> for minimization says: if $x_*\in X$ satisfies $h(x_*)=b$ and globally minimizes $L_b(\,\cdot\,,\lambda_*)$ on $X$ for some $\lambda_*$, then it globally minimizes the constrained <objective function>. Indeed, for every feasible $x$,
$$
f(x)=L_b(x,\lambda_*)\ge L_b(x_*,\lambda_*)=f(x_*).
$$
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 $\phi(u)=\inf\{f(x):x\in X,\ h(x)=u\}$, setting $\inf\varnothing=+\infty$, and assume $\phi(b)$ is finite. The <Strong Lagrangian property> at $b$ means that a finite multiplier attains the constrained value as an unconstrained lower bound:
$$
\boxed{\inf_{x\in X}L_b(x,\lambda_*)=\phi(b).}
$$
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 $\lambda_*$ and supports the <epigraph> from below:
$$
\phi(u)\ge\phi(b)+\lambda_*^T(u-b)\quad\text{for every }u.
$$
If this inequality holds, then for every $x\in X$,
$$
f(x)\ge\phi(h(x))\ge\phi(b)+\lambda_*^T(h(x)-b),
$$
so $\inf_XL_b(x,\lambda_*)\ge\phi(b)$. Conversely restricting the infimum to $h(x)=b$ gives $\inf_XL_b\le\phi(b)$, even if the primal infimum is approached only by a sequence. Thus equality holds. In the reverse direction, the Strong <optimization Lagrangian> identity implies $f(x)-\lambda_*^T(h(x)-b)\ge\phi(b)$ 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 $\phi(b)$.
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 $x,y\ge0$. Substitute $y=B-3x$ to reduce the <maximization problem> to
$$
F_B(x)=-14x^2+(7B-1)x+3B-B^2,\qquad 0\le x\le B/3.
$$
Its <second derivative> is $-28$, so the unique constrained maximum is obtained by clipping the <stationary point>:
$$
\boxed{(x_*,y_*)=
\begin{cases}
(0,B),&0\le B\le1/7,\\
(B/4-1/28,\ B/4+3/28),&B\ge1/7.
\end{cases}}
$$
The upper constraint never clips the interior formula because $B/4-1/28<B/3$ for $B\ge0$. The maximal revenue in thousands of pounds is
$$
R(B)=
\begin{cases}
3B-B^2,&0\le B\le1/7,\\
-B^2/8+11B/4+1/56,&B\ge1/7.
\end{cases}
$$
Equivalently the revenue <Hessian> is $\begin{pmatrix}-4&1\\1&-2\end{pmatrix}$, <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,
$$
\begin{array}{c|c|c|c}
a&x_*&y_*&R(a/1000)\\ \hline
1000&3/14&5/14&37/14\\
10000&69/28&73/28&841/56
\end{array}
$$
The units in the middle columns are minutes; the last column is thousands of pounds. Revenue measured in pounds is $G(a)=1000R(a/1000)$, so
$$
\boxed{G'(1000)=\frac52,\qquad G'(10000)=\frac14.}
$$
An 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 $G(a+1)-G(a)=G'(a)-1/8000$.
The maximization multiplier in $f(x,y)-\mu(3x+y-B)$ equals $R'(B)=(11-B)/4$ 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.
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