带约束的$f(x,y,z)=x+y+z$全局最小值存在性及最大值判定

Hey folks, let's break down this problem into its two core tasks and work through each step carefully.

Task (ii): Does $f$ Have a Maximum Value? #

Short answer: No, $f$ does not have a maximum under the given constraint. Here's how to prove it with a simple construction:

• Set $y=0$: the constraint simplifies to $x^2 - z = 1$, so $z = x^2 - 1$.
• Substitute into $f$: $f(x, 0, x^2-1) = x + 0 + (x^2 - 1) = x^2 + x - 1$.
• As $x \to +\infty$, the quadratic term $x^2$ dominates, so $x^2 + x -1$ tends to $+\infty$. This means we can make $f$ arbitrarily large by choosing larger and larger positive $x$ (with $y=0$ and $z$ set to satisfy the constraint), so no maximum exists.

Task (i): Finding the Global Minimum Value #

To find the minimum, we first need to confirm it exists, then use Lagrange multipliers to compute it, and finally verify it's the global minimum.

Step 1: Proving a Global Minimum Exists #

We can use the properties of continuous functions and quadratic forms here:

1. From the constraint, substitute $z = x^2 + xy + 2y^2 -1$ into $f$, giving us a function of just $x$ and $y$:
   $$f(x,y) = x + y + x^2 + xy + 2y^2 -1$$
2. Look at the quadratic term: $x^2 + xy + 2y^2$. The corresponding matrix for this quadratic form is $\begin{pmatrix}1 & 0.5 \ 0.5 & 2\end{pmatrix}$. Its determinant is $1*2 - (0.5)^2 = 1.75 > 0$, and the leading principal minors are positive, so this is a positive definite quadratic form.
3. Positive definite quadratic forms grow to $+\infty$ as $|(x,y)| \to +\infty$, and the linear terms ($x + y$) grow much slower. This means $f(x,y)$ tends to $+\infty$ as $x$ and/or $y$ get arbitrarily large.
4. Since $f$ is continuous and tends to $+\infty$ at "infinity", it must have a lower bound on the constraint set, and by the extreme value theorem (applied to a sufficiently large closed ball where $f$ is smaller than any value outside), $f$ attains its global minimum somewhere on the constraint set.

Step 2: Calculating the Minimum with Lagrange Multipliers #

Construct the Lagrangian function:
$$\mathcal{L}(x,y,z,\lambda) = x + y + z - \lambda\left(x^2 + xy + 2y^2 - z - 1\right)$$

Take partial derivatives and set them to zero:

• $\frac{\partial \mathcal{L}}{\partial x} = 1 - \lambda(2x + y) = 0$ → $2x + y = \frac{1}{\lambda}$
• $\frac{\partial \mathcal{L}}{\partial y} = 1 - \lambda(x + 4y) = 0$ → $x + 4y = \frac{1}{\lambda}$
• $\frac{\partial \mathcal{L}}{\partial z} = 1 + \lambda = 0$ → $\lambda = -1$
• $\frac{\partial \mathcal{L}}{\partial \lambda} = -\left(x^2 + xy + 2y^2 - z - 1\right) = 0$ → $z = x^2 + xy + 2y^2 -1$

Substitute $\lambda = -1$ into the first two equations to get a linear system:
$$
\begin{cases}
2x + y = -1 \
x + 4y = -1
\end{cases}
$$

Solve this system:

• Multiply the first equation by 4: $8x + 4y = -4$
• Subtract the second equation: $7x = -3$ → $x = -\frac{3}{7}$
• Plug $x = -\frac{3}{7}$ back into $2x + y = -1$: $y = -1 - 2*(-\frac{3}{7}) = -\frac{1}{7}$

Now compute $z$ using the constraint:
$$z = \left(-\frac{3}{7}\right)^2 + \left(-\frac{3}{7}\right)\left(-\frac{1}{7}\right) + 2\left(-\frac{1}{7}\right)^2 -1 = \frac{9 + 3 + 2}{49} -1 = \frac{14}{49} -1 = -\frac{5}{7}$$

Finally, calculate $f$ at this point:
$$f\left(-\frac{3}{7}, -\frac{1}{7}, -\frac{5}{7}\right) = -\frac{3}{7} - \frac{1}{7} - \frac{5}{7} = -\frac{9}{7}$$

Step 3: Confirming This is the Global Minimum #

Since we proved a global minimum exists, and the Lagrange multiplier method gave us exactly one critical point, plus the quadratic form is positive definite (meaning the function is convex in $x$ and $y$), this critical point must be the global minimum.

内容的提问来源于stack exchange，提问作者chrstnsn