Find infimum and supremum of the function $f$ defined by the formula $f(x, y, z) = x^2 + y^2 + z^2$ on the set $A$

Question

Let $A = \{(x, y, z) \in \mathbb R^3 : 5x^2 + 5y^2 - z^2 = 0, x + 2y +3z = 20\}$. Find infimum and supremum of the function $f$ defined by the formula $f(x, y, z) = x^2 + y^2 + z^2$ on the set $A$.

Using Lagrange multipliers I found critical points $(1,2,5)$, $(-2, -4, 10)$. However, I don't know how to check whether these are infimum and supremum of the function or just local extremes. It would be helpful if the set $A$ was compact - then the found extremes must be the infimum and supremum of the $f$ function, but I have no idea how to check the compactness of $A$.

Answer 1

Intersecting a conical surface with a plane surface of Cartesian equations:

$$ x^2 + y^2 = k^2z^2, \quad \quad \quad a\,x + b\,y + c\,z + d = 0 $$

after calculating the discriminant:

$$ \Delta := c^2-(a^2+b^2)k^2 $$

• if $d = 0$ we can find three degenerate cases:

  • if $\Delta<0$ then they're two straight lines;

  • if $\Delta=0$ then it's a straight line;

  • if $\Delta>0$ then it's a point;

• if $d \ne 0$ we can find three non-degenerate cases:

  • if $\Delta<0$ then they're two branches of hyperbola;

  • if $\Delta=0$ then it's a parabola;

  • if $\Delta>0$ then it's an ellipse.

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In the present case:

$$ x^2 + y^2 = \frac{1}{5}z^2, \quad \quad \quad x + 2\,y + 3\,z - 20 = 0 $$

then:

$$ d = -20 \ne 0, \quad \quad \quad \Delta = 3^2-\left(1^2+2^2\right)\frac{1}{5} = 8 > 0 $$

so it's an ellipse.

Therefore, since $f(x,y,z)=x^2+y^2+z^2$ is a continuous function on a closed and bounded set, there it admits a global minimum $f(1,2,5)=30$ and a global maximum $f(-2,-4,10)=120$.

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If we were also interested in a parametrization of this ellipse, from the plane equation:

$$ z = \frac{20-x-2\,y}{3} $$

which substituted into the cone equation leads to:

$$ 44\,x^2 - 4\,x\,y + 41\,y^2 + 40\,x + 80\,y - 400 = 0 $$

which through a study of the eigenvalues/eigenvectors of this conic leads to:

$$ \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} \\ -1 \\ \end{bmatrix} + \begin{bmatrix} -\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ -\frac{2}{\sqrt{5}} & -\frac{1}{\sqrt{5}} \\ \end{bmatrix} \begin{bmatrix} \frac{3\sqrt{5}}{2}\,\cos\theta \\ \sqrt{10}\,\sin\theta \\ \end{bmatrix} $$

i.e.

$$ \begin{cases} x = -\frac{1}{2} - \frac{3}{2}\,\cos\theta + 2\sqrt{2}\,\sin\theta \\ y = -1 -3\cos\theta - \sqrt{2}\,\sin\theta \\ z = \frac{15}{2} + \frac{5}{2}\,\cos\theta \\ \end{cases} \quad \quad \text{with} \; \theta \in [0,2\pi). $$

Of course, for the purposes of this exercise, it isn't strictly necessary.