Minimize a quadratic function of three variables over a plane

Question

I have the following minimization problem:

If $ x + 2 y + 4 z = 10 $, what is the minimum of

$ f(x,y,z) = x^2 + 2 y^2 + 3 z^2 + x y + 2 xz + 3 y z + 7 z $

and at what value of $(x,y,z)$ does the minimum occur ?

My approach:

Define the Lagrange function

$ g(x,y,z) = f(x,y,z) + \lambda (x + 2 y + 4 z - 10 ) $

Then it follows by taking partial derivatives with respect to $x, y, z, \lambda$, that

$ 2 x+ y + 2 z + \lambda = 0$

$ 4 y + x + 3 z + 2 \lambda = 0 $

$ 6 z + 2 x + 3 y + 7 + 4 \lambda = 0 $

$ x + 2 y + 4 z = 10 $

This is a linear system in $x, y, z, \lambda$, and is readily solvable. The solution is

$ (x,y,z, \lambda ) = (0.\overline{36} , 1.4, 1.7\overline{09}, -5.\overline{54} ) $

And the value of the minimum of $f$ is

$ f_{Min} = 33.7\overline{09} $

My question is: How do I know that this is actually a minimum (the global minimum), and not a saddle point ? Is there a test I can perform on the results (particularly the value of $\lambda$) that guarantees that this is indeed the global minimum ? Or, do I have to find the Hessian matrix, by differentiating all the above partial derivatives again, to find the second order partial derivatives ?

I've computed the Hessian, it is given by

$ H = \begin{bmatrix} 2 && 1 && 2 && 1 \\ 1 && 4 && 3 && 2 \\ 2 && 3 && 6 && 4 \\ 1 && 2 && 4 && 0 \end{bmatrix} $

But $H$ is indefinite (i.e it has positive as well as negative eigenvalues). However the $3 \times 3$ principal submatrix defined below

$ H_{3 \times 3} = \begin{bmatrix} 2 && 1 && 2 \\ 1 && 4 && 3 \\ 2 && 3 && 6 \end{bmatrix} $

is positive definite. So is this what I should be looking for ?

Answer 1

$$f(x,y,z)\to x^2+x y+2 x z+2 y^2+3 y z+3 z^2+7 z$$

With $z=\frac{1}{4} (-x-2 y+10)$ we define

$$g(x,y)=f(x,y,\frac{1}{4} (-x-2 y+10))=\frac{1}{16} \left(11 x^2-8 x+20 y^2-56 y+580\right)$$

We get the gradient of $g(x,y)$:

$$\nabla _{\{x,y\}}g(x,y)=\left\{\frac{1}{16} (22 x-8),\frac{1}{16} (40 y-56)\right\}=0$$

Solving the equation we obtain $\left\{x\to \frac{4}{11},y\to \frac{7}{5}\right\}$ and $z\to\frac{1}{4} (-x-2 y+10)=\frac{94}{55}$

Now we want the Hessian matrix:

$$H=\text{Hessian}(g(x,y))=\left( \begin{array}{cc} \frac{11}{8} & 0 \\ 0 & \frac{5}{2} \\ \end{array} \right)$$

with determinant $| H|=\frac{55}{16}>0$ the point $\left\{x\to \frac{4}{11},y\to \frac{7}{5},z\to \frac{94}{55}\right\}$ is actually a minimum.

Hence $f_{min}(\frac{4}{11},\frac{7}{5},\frac{94}{55})=g_{min}\left(\frac{4}{11},\frac{7}{5}\right)=\frac{1854}{55}$ is the value of the minimum.

Answer 2

Once $f(x,y,z):=g(x,y)$ has been calculated as $$g(x,y)=ax^2+bx+ cy^2+dx + e$$ with $a,c>0,$ the minimum is attained at $$x=-{b\over 2a},\quad y=-{c\over 2d}$$ No differential calculus is needed, just a simple formula for the quadratic function of one variable.