Constrained Minimum via Lagrange Multipliers: 04-BS-1, December 2019, Q3

Question 3: 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 $F=2x^2+y^2+3z^2$ (positive-definite quadratic form); constraint $g=x+y-z+1=0$ (a plane).

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

Approach. Set $\nabla F=\lambda\nabla g$ componentwise, solve for $x,y,z$ in terms of $\lambda$, substitute into the constraint to find $\lambda$, then evaluate $F$. Since $F$ is a positive-definite quadratic form and the constraint is an unbounded plane, the unique critical point found this way is automatically the global minimum.

Lagrange equations. $\nabla F=(4x,2y,6z)$, $\nabla g=(1,1,-1)$, giving $$4x=\lambda,\qquad 2y=\lambda,\qquad 6z=-\lambda\ \Longrightarrow\ x=\tfrac\lambda4,\ y=\tfrac\lambda2,\ z=-\tfrac\lambda6.$$
Substitute into the constraint. $$\tfrac\lambda4+\tfrac\lambda2-\left(-\tfrac\lambda6\right)+1=0\ \Longrightarrow\ \tfrac{3\lambda+6\lambda+2\lambda}{12}=-1\ \Longrightarrow\ \tfrac{11\lambda}{12}=-1\ \Longrightarrow\ \lambda=-\tfrac{12}{11}.$$
Recover $x,y,z$. $$x=-\tfrac3{11},\qquad y=-\tfrac6{11},\qquad z=\tfrac2{11}.$$
Evaluate $F$. $$F=2\!\left(\tfrac3{11}\right)^{\!2}+\left(\tfrac6{11}\right)^{\!2}+3\!\left(\tfrac2{11}\right)^{\!2}=\frac{18+36+12}{121}=\frac{66}{121}=\frac6{11}.$$ Since $F$ is positive-definite and the constraint plane is unbounded, this unique critical point is the global minimum (no maximum exists on an unbounded plane).

$$F_{\min}=\boxed{\tfrac6{11}}\ \text{at}\ \left(-\tfrac3{11},\,-\tfrac6{11},\,\tfrac2{11}\right)$$

Quantity	Result
$\lambda$	$-\tfrac{12}{11}$
Critical point	$(-\tfrac3{11},-\tfrac6{11},\tfrac2{11})$
$F_{\min}$	$\tfrac6{11}$

Concept & Theory

Topic. Lagrange multipliers for a quadratic objective constrained to a plane.

Key relations. At a constrained extremum, $\nabla F=\lambda\nabla g$; combined with $g=0$ this pins down a single critical point when the constraint is linear and $F$ is a positive-definite quadratic form.

Why this works. A positive-definite quadratic form restricted to any plane is itself a (shifted, lower-dimensional) positive-definite quadratic form, which has a unique global minimum and no maximum — so the single critical point from the Lagrange system must be that minimum, without needing to compare multiple candidates as one would on a compact surface.

Common pitfall. Assuming a maximum also exists (it does not — the plane is unbounded and $F\to\infty$ as $\|(x,y,z)\|\to\infty$ along it); arithmetic slips combining the three fractions with denominator $12$ before solving for $\lambda$.

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

Question 3 of 8: Constrained Minimum via Lagrange Multipliers

Question 3: Constrained Minimum via Lagrange Multipliers (20 marks)