What is the minimum value of $x+4z$, a function defined on $\mathbb{R^3}$, subject to the constraint $x^2 + y^2 +z^2 \leq 2$?
I know how to solve this if the constraint is an equality, but what shall I do if it's an inequality? Could anyone help me, please?
On
You might as well force $y=0$ since that gives the largest set of pairs $(x,z)$.
Thus, the new problem is to minimize $x+4z$, defined on $\{(x,z) \in \mathbb{R^2}\}$, subject to the constraint $x^2 + z^2 \leq 2$.
Clearly, the minimum value of $x+4z$ will be negative, and the point $(x,z)$ where the minimum occurs will have $x,z$ both non-positive (with at least one of$\;x,z\;$negative).
But then, for a point $(x,z)$ where the minimum occurs, scaling the point $(x,z)$ by a constant greater than $1$ would make $x+4z$ even more negative.
It follows that the minimum value of $x+4z$ will occur on the part of the circle $x^2 + z^2=2$ in the third quadrant (possibly including the endpoints of the arc).
So now you have minimization with an equality constraint.
Can you finish it?
On
We have the following quadratically constrained linear program (QCLP) in $\mathrm x \in \mathbb R^3$
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm x\\ \text{subject to} & \| \mathrm x \|_2^2 \leq 2\end{array}$$
where $\mathrm c := (1,0,4)$. Since the objective function is linear and nonzero, the minimum is attained at the boundary, i.e., on the Euclidean sphere of radius $\sqrt 2$ and centered at the origin. Hence, we have the following equality-constrained QCLP in $\mathrm x \in \mathbb R^3$
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm x\\ \text{subject to} & \| \mathrm x \|_2^2 = 2\end{array}$$
Using the Cauchy-Schwarz inequality,
$$\mathrm c^\top \mathrm x \geq - \| \mathrm c \|_2 \| \mathrm x \|_2 = - \sqrt{17} \cdot \sqrt{2} = - \sqrt{34}$$
Thus, the minimum is $\color{blue}{-\sqrt{34}}$, which is attained at $\mathrm x_{\min} = \gamma \mathrm c$, where $|\gamma| = \frac{\sqrt{2}}{\| \,\mathrm c \|_2} = \sqrt{\frac{2}{17}}$. Thus,
$$\mathrm x_{\min} = \color{blue}{- \begin{bmatrix} \sqrt{\frac{2}{17}}\\ 0\\ 4 \sqrt{\frac{2}{17}}\end{bmatrix}}$$
Let the Lagrangian be
$$\mathcal L (\mathrm x, \lambda) := \mathrm c^\top \mathrm x + \lambda \left( \| \mathrm x \|_2^2 - 2 \right)$$
Taking the partial derivatives of $\mathcal L$ and finding where they vanish, we obtain
$$\begin{array}{rl} \mathrm c + 2 \lambda \mathrm x &= 0_3\\ \| \mathrm x \|_2^2 &= 2\end{array}$$
Hence, $\mathrm x = - \frac{1}{2 \lambda} \mathrm c$ and, from $\| \mathrm x \|_2^2 = 2$, we obtain $\lambda = \pm \sqrt{\frac{17}{8}}$. Thus,
$$\mathrm x_{\min} = - \frac{1}{2 \lambda} \mathrm c = - \sqrt{\frac{2}{17}} \, \mathrm c = \color{blue}{- \begin{bmatrix} \sqrt{\frac{2}{17}}\\ 0\\ 4 \sqrt{\frac{2}{17}}\end{bmatrix}}$$
and the minimum is
$$\mathrm c^\top \mathrm x_{\min} = - \sqrt{\frac{2}{17}} \| \mathrm c \|_2^2 = \color{blue}{-\sqrt{34}}$$
Using the Schur complement test for positive semidefiniteness, the original inequality-constrained QCLP can be rewritten as the following semidefinite program (SDP)
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm x\\ \text{subject to} & \begin{bmatrix} \mathrm I_3 & \mathrm x \\ \mathrm x^{\top} & 2\end{bmatrix} \succeq \mathrm O_4\end{array}$$
The objective function $u=x+4z$ does not depend on $y$, therefore $y$ must be minimum, that is $y=0$ so that $x,z$ are the maximum negative values.
Thus the new optimization problem states: Minimize $u=x+4z$ subject to $x^2+z^2\le 2.$
The feasible region is a circle with the radius $\sqrt{2}$ and the contour line of the objective function is $z=-\frac14x+\frac14u$.
Minimum of $u(x,z)$ occurs when the contour line is tangent to the feasible region from below: $$z=-\sqrt{2-x^2} \Rightarrow z'=\frac{x}{\sqrt{2-x^2}}=-\frac14 \Rightarrow x=-\sqrt{\frac{2}{17}} \Rightarrow z=-\sqrt{\frac{32}{17}} \Rightarrow$$ $$u=-\sqrt{\frac{2}{17}}-4\sqrt{\frac{32}{17}}=-\sqrt{34} \ (min).$$