Question

Find the minimum and maximum of f(x,y,z)=x2+y2+z2 subject to two constraints, x+2y+z=7 and x−y=6.

Answer 1

Use the method of Lagrange multipliers. We have Lagrangian

$L(x,y,z,\lambda_1,\lambda_2)=x^2+y^2+z^2+\lambda_1(x+2y+z-7)+\lambda_2(x-y-6)$

with partial derivatives (set equal to 0) of

$L_x=2x+\lambda_1+\lambda_2=0$
$L_y=2y+2\lambda_1-\lambda_2=0$
$L_z=2z+\lambda_1=0$
$L_{\lambda_1}=x+2y+z-7=0$
$L_{\lambda_2}=x-y-6=0$

As $x+2y+z=7$, and $x-y=6$, we can obtain

$\dfrac12L_x+L_y+\dfrac12L_z=0\implies3\lambda_1-\dfrac12\lambda_2=-7$
$L_x-L_y=0\implies\lambda_1-2\lambda_2=12$
$\begin{cases}3\lambda_1-\frac12\lambda_2=-7\\\lambda_1-2\lambda_2=12\end{cases}\implies\lambda_1=-\dfrac{40}{11},\lambda_2=-\dfrac{86}{11}$

From this, we find a single critical point:

$2x-\dfrac{40}{11}-\dfrac{86}{11}=0\implies x=\dfrac{63}{11}$
$\dfrac{63}{11}-y=6\implies y=-\dfrac3{11}$
$\dfrac{63}{11}-\dfrac6{11}+z=7\implies z=\dfrac{20}{11}$

At this point, we have a value of

$f\left(\dfrac{63}{11},-\dfrac3{11},\dfrac{20}{11}\right)=\dfrac{398}{11}$

To determine what kind of extremum occurs at this point, we check the Hessian of $f(x,y,z)$:

$\mathbf H(x,y,z)=\begin{bmatrix}f_{xx}&f_{xy}&f_{xz}\\f_{yx}&f_{yy}&f_{yz}\\f_{zx}&f_{zy}&f_{zz}\end{bmatrix}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}$

We observe that $\det\mathbf H(x,y,z)=8>0$ at any point $(x,y,z)$, and that the eigenvalues of this matrix are all positive (2 with multiplicity 3), so $\mathbf H$ is positive definite. By the second partial derivative test, this means $f(x,y,z)$ attains a minimum at this critical point. Meanwhile, $f$ has no maximum value.