Constrained Minimum via Lagrange Multipliers: 04-BS-1, May 2016, Q4

Question 4: Constrained Minimum via Lagrange Multipliers 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Objective $F(x,y,z)=2x^2+y^2+3z^2$ (a positive-definite quadratic form, so it has a global minimum on any plane) and constraint $g(x,y,z)=x+y-z+1=0$.

Find. The minimum value of $F$ on the constraint plane.

Approach. Use Lagrange multipliers: $\nabla F=\lambda\nabla g$, combined with the constraint equation, gives four equations in $x,y,z,\lambda$.

Set up the Lagrange conditions. $\nabla F=(4x,2y,6z)$ and $\nabla g=(1,1,-1)$, so $$4x=\lambda,\qquad 2y=\lambda,\qquad 6z=-\lambda \;\Rightarrow\; x=\frac{\lambda}{4},\ \ y=\frac{\lambda}{2},\ \ z=-\frac{\lambda}{6}.$$
Substitute into the constraint. $$\frac{\lambda}{4}+\frac{\lambda}{2}-\left(-\frac{\lambda}{6}\right)+1=0 \;\Rightarrow\; \frac{3\lambda+6\lambda+2\lambda}{12}=-1 \;\Rightarrow\; \frac{11\lambda}{12}=-1 \;\Rightarrow\; \lambda=-\frac{12}{11}.$$
Recover the critical point. $$x=\frac{-12/11}{4}=-\frac{3}{11},\qquad y=\frac{-12/11}{2}=-\frac{6}{11},\qquad z=-\frac{-12/11}{6}=\frac{2}{11}.$$ Check: $x+y-z+1=-\tfrac{3}{11}-\tfrac{6}{11}-\tfrac{2}{11}+1=-\tfrac{11}{11}+1=0$ ✓.
Evaluate $F$ at the critical point. Since $F$ is a positive-definite quadratic form (sum of positive squares) and the constraint is an unbounded plane, $F\to\infty$ as $\|(x,y,z)\|\to\infty$ along the plane, so this unique critical point must be the global minimum: $$F=2\left(\frac{3}{11}\right)^2+\left(\frac{6}{11}\right)^2+3\left(\frac{2}{11}\right)^2=\frac{18+36+12}{121}=\frac{66}{121}=\boxed{\dfrac{6}{11}}$$

Quantity	Result
Critical point	$(x,y,z)=\left(-\tfrac{3}{11},-\tfrac{6}{11},\tfrac{2}{11}\right)$
Multiplier $\lambda$	$-\tfrac{12}{11}$
Minimum value $F_{\min}$	$\dfrac{6}{11}\approx0.545$

Concept & Theory

Topic. Constrained optimization via Lagrange multipliers.

Key relations. At a constrained extremum of $F$ subject to $g=0$, $\nabla F=\lambda\nabla g$ for some scalar $\lambda$ — the objective's gradient must be parallel to the constraint surface's normal, since any component of $\nabla F$ tangent to the constraint could otherwise be used to further decrease (or increase) $F$ while staying on the surface.

Why this works. Because $F$ is a positive-definite quadratic form (a sum of positive multiples of squares), its Lagrange system has exactly one critical point, and the sublevel sets $\{F\le k\}$ are bounded ellipsoids that grow without bound as $k$ increases. They must therefore eventually intersect the unbounded constraint plane, meaning $F$ restricted to the plane attains a global minimum somewhere — and by the uniqueness of the Lagrange critical point, it is exactly the point found here, not merely a saddle point of the constrained problem.

Common pitfall. Sign errors in $\nabla g=(1,1,-1)$, particularly the $-1$ contributed by the $-z$ term inside the constraint $x+y-z+1=0$. Also, reporting the stationary value of $F$ without justifying why it must be a minimum (rather than assuming this without the boundedness/positive-definiteness argument above) — on an unbounded constraint set, a single critical point is not automatically an extremum unless the objective's structure guarantees it.

Source. Stewart, Calculus: Early Transcendentals, Ch. 14 (§14.8 Lagrange Multipliers).

Question 4 of 8: Constrained Minimum via Lagrange Multipliers

Question 4: Constrained Minimum via Lagrange Multipliers 20 marks