Question bank — Isoperimetric problems — constraints (Lagrange multipliers in variational sense)

False. Plain E–L wanders over all admissible functions, including those that break $K=\ell$; you must optimise the augmented $F=f-\lambda g$ so variations stay on the constraint surface.

True. An integral constraint contributes one number's worth of restriction, so one constant $\lambda$ suffices — contrast with pointwise constraints $g(x,y)=0$ which need $\lambda (x)$.

False. $\lambda$ is fixed by the constraint; change $\ell$ and $\lambda$ generally changes too — indeed $\lambda =d{J}^{\ast }/d\ell$ is the marginal rate of change (see Shadow Price / Envelope Theorem).

False. One knob cannot both keep $K$ fixed and probe stationarity of $J$; you need the two-parameter family $\bar{y}+{ϵ}_1{\eta }_1+{ϵ}_2{\eta }_2$ for the multiplier to emerge.

True. Integrating the Beltrami first integral gives $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2$, so $\lambda$ literally is the circle's radius.

False. Minimising height of centre of mass at fixed length gives $y=\lambda +C\cosh \left(\frac{x-x_0}{C}\right)$ — a hyperbolic cosine (Catenary Curve), which only looks parabolic near the bottom.

True. There are three unknowns $C_1,C_2,\lambda$, matched by three conditions: the two boundary values $y(a),y(b)$ plus the constraint $\int gdx=\ell$.

The E–L equation of $F$ is unchanged in form by the sign; the sign only flips $\lambda$'s numerical value. But if you want $\lambda$ to carry its shadow-price meaning $d{J}^{\ast }/d\ell$, be consistent — the sign is not "wrong", just a bookkeeping convention you must not mix mid-problem.

Error: ${\eta }_1$ is spent enforcing $\delta K=0$, so it is not free. Only ${\eta }_2$ remains essentially arbitrary, and it is that one degree of freedom that the fundamental lemma acts on to force the bracket to vanish.

Error: because boundary values are fixed, every bump satisfies ${\eta }_i(a)={\eta }_i(b)=0$, so those boundary terms are exactly zero. They vanish, not contribute.

Error: $\lambda$ is determined by the constraint ${\int }_a^{b}gdx=\ell$, not by re-optimising $J$. Plug the solved $y(x)$ (still containing $\lambda$) into $K$ and set it equal to $\ell$.

Error: Beltrami Identity is a first integral only when $F=f-\lambda g$ has no explicit $x$. If $f$ or $g$ depends on $x$ directly, you must solve the full E–L equation.

Error: length is the constraint here, not the objective. Dido maximises area at fixed length; the straight line is the shortest curve, giving zero enclosed area — the exact opposite extremal.

Deleting it changes the problem; adding $\lambda \times$(constraint) leaves the optimum unchanged on the surface $K=\ell$ (since the added term is a fixed constant there) while converting a hard constrained search into a free one.

Because the two-parameter family collapses the functional problem into an ordinary constrained extremum in $({ϵ}_1,{ϵ}_2)$-space, where the finite-dimensional rule directly applies; the variational statement is the limit of that.

By the envelope theorem, $\lambda =d{J}^{\ast }/d\ell$: it tells you how much the optimal objective improves per unit of relaxed budget — the marginal value of loosening the constraint.

An integral constraint imposes one global scalar condition, matched by one scalar $\lambda$; a pointwise condition holds at every $x$, so it needs a whole function $\lambda (x)$ to enforce it everywhere.

The augmented Beltrami integral is solvable in closed form and yields $(x-x_0{)}^2+(y-C{)}^2={\lambda }^2$ — the only curve of constant curvature — which is the equality case of the Isoperimetric Inequality.

Beltrami is a first integral: it has already integrated the E–L equation once using the absence of explicit $x$, trading one order of differentiation for one integration constant $C$.

Then $\lambda =0$ and the constraint is inactive — the isoperimetric solution coincides with the unconstrained one, just as a slack Lagrange constraint gives a zero multiplier.

The only curve of that length joining them is the straight segment, so the "search" is degenerate — there is a unique admissible function and no room to optimise; the extremal is forced.

No admissible curve exists — you cannot connect two points with a shorter arc than the straight line, so the constrained problem is empty and has no solution.

The curve tightens toward the straight chord; in the $\cosh$ solution $C\to \infty$, flattening the hyperbolic cosine into a nearly straight line (its parabola-like bottom becomes the whole visible curve).

Replace the constant $\lambda$ by a multiplier function $\lambda (x)$ and form $F=f-\lambda (x)g$; the constraint must then hold at every $x$, not just on average.

It returns no interior extremal; the optimum then lies on a boundary of the admissible set (e.g. an endpoint configuration), which the E–L-plus-multiplier machinery cannot see — it only finds interior stationary curves.