Maxima, minima, and Lagrange multipliers

Unconstrained critical points. If $f(x,y)$ has a local max or min at an interior point $(a,b)$ where $f$ is differentiable, both partial derivatives must vanish there:

where: $f_x,f_y$ are the first partial derivatives of $f$. Points satisfying this are critical points, but the condition is only necessary — a critical point can be a max, a min, or (new to several variables) a saddle.

Classifying with the discriminant. At a critical point $(a,b)$, define

where: $f_{xx},f_{yy},f_{xy}$ are second partial derivatives evaluated at the critical point. If $D>0$ and $f_{xx}>0$, it is a local minimum; if $D>0$ and $f_{xx}<0$, a local maximum; if $D<0$, a saddle point; if $D=0$, the test is inconclusive. This test is really a compact restatement of the second-order Taylor expansion of $f$ near $(a,b)$: the quadratic terms $\tfrac{1}{2}\big[f_{xx}h^2+2f_{xy}hk+f_{yy}k^2\big]$ (for a small step $(h,k)$ away from the critical point) keep a fixed sign in every direction exactly when $D>0$, which is what makes the point a clean max or min; when $D<0$ the quadratic form changes sign with direction, producing the mixed up-and-down curvature of a saddle. A quick illustration: for $f(x,y)=x^3-3xy+y^3$, solving $f_x=3x^2-3y=0$ and $f_y=-3x+3y^2=0$ gives critical points $(0,0)$ and $(1,1)$. Since $f_{xx}=6x$, $f_{yy}=6y$, $f_{xy}=-3$, at $(0,0)$ we get $D=(0)(0)-(-3)^2=-9<0$, a saddle point, while at $(1,1)$ we get $D=(6)(6)-(-3)^2=27>0$ with $f_{xx}=6>0$, a local minimum of value $f(1,1)=1-3+1=-1$.

Constrained extrema. To extremize $f(x,y,z)$ subject to a single equality constraint $g(x,y,z)=0$, introduce a multiplier $\lambda$ and solve

where: $\nabla f=(f_x,f_y,f_z)$ and $\nabla g=(g_x,g_y,g_z)$ are the gradients, and $\lambda$ is the Lagrange multiplier — an unknown scalar solved for alongside $x,y,z$. This is $n+1$ equations in $n+1$ unknowns for $n$ variables plus the constraint, and it handles only equality constraints; inequality-constrained (KKT) problems are beyond this course.

Use Lagrange multipliers to find the point on the plane $x+2y+3z=6$ closest to the origin, and hence the minimum distance.

Step 1. Minimizing distance is equivalent to (and algebraically easier via) minimizing the squared distance $f(x,y,z)=x^2+y^2+z^2$, subject to $g(x,y,z)=x+2y+3z-6=0$.

Step 2. Form the Lagrange equations $\nabla f=\lambda\nabla g$ component by component: $f_x=2x=\lambda(1)$, $f_y=2y=\lambda(2)$, $f_z=2z=\lambda(3)$.

Step 3. Solve for $x,y,z$ in terms of $\lambda$: $x=\dfrac{\lambda}{2}$, $y=\lambda$, $z=\dfrac{3\lambda}{2}$.

Step 4. Substitute into the constraint $x+2y+3z=6$: $\dfrac{\lambda}{2}+2\lambda+\dfrac{9\lambda}{2}=6$, which collects to $7\lambda=6$, so $\lambda=\dfrac{6}{7}$.

Step 5. Back-substitute to get the closest point: $x=\dfrac{3}{7}$, $y=\dfrac{6}{7}$, $z=\dfrac{9}{7}$.

Step 6. Compute the distance. $f(x,y,z)=\dfrac{9}{49}+\dfrac{36}{49}+\dfrac{81}{49}=\dfrac{126}{49}=\dfrac{18}{7}$ is the minimum squared distance, so the minimum distance itself is

Competitive stretch

Check the answer with the point-to-plane distance shortcut, $d=\dfrac{|Ax_0+By_0+Cz_0-D|}{\sqrt{A^2+B^2+C^2}}$ for a plane $Ax+By+Cz=D$ and point $(x_0,y_0,z_0)$. Here $(x_0,y_0,z_0)=(0,0,0)$, so $d=\dfrac{|0+0+0-6|}{\sqrt{1+4+9}}=\dfrac{6}{\sqrt{14}}=\dfrac{6\sqrt{14}}{14}=\dfrac{3\sqrt{14}}{7}$, matching Step 6 exactly and confirming that Lagrange's method reproduces the standard geometry formula.

• Forgetting the companion check on $f_{xx}$. $D>0$ alone only says "extremum," not which kind — the sign of $f_{xx}$ (equivalently $f_{yy}$, which shares the same sign when $D>0$) decides max versus min.
• Reading $D=0$ as "no extremum" instead of "inconclusive." The discriminant test is silent at $D=0$; a direct check or higher-order terms is needed.
• Dropping the final square root. Minimizing $x^2+y^2+z^2$ gives the same location as minimizing $\sqrt{x^2+y^2+z^2}$, but the minimized value is the squared distance — forgetting to take the square root at the end is a common slip.
• Mixing up which gradient is scaled by $\lambda$. Each component equation $f_{x_i}=\lambda g_{x_i}$ must be set up consistently, and the constraint equation itself is always the extra equation needed to pin down $\lambda$.

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