## 数学代写|多变量微积分代写multivariable calculus代考|Optimization over compact domains

Recall our definition of a compact set (Definition 1.8): a set $\Omega \subseteq \mathbb{R}^n$ is said to be compact if it is closed and bounded.

For any function defined on a region $\Omega \subseteq D_f$ that is compact, we have the following very useful result.

Firstly, it is not necessary that the region being considered is the function’s entire domain of definition, $D_f$, but it might be. The problem statement will usually specify this. If no region is given then the reader should assume the whole of $D_f$ is implied.

Secondly, by Theorem 1.2, a continuous function defined on a closed and bounded region is necessarily bounded. This means that $|f(\boldsymbol{x})|<K$ for some $K \in \mathbb{R}$ and for all $\boldsymbol{x}$ in that region. This simple result implies that we should expect $f$ to exhibit an absolute minimum and an absolute maximum. In fact, this is the only time we are guaranteed that absolute maximum and minimum points exist.

The reader should always bear in mind that a continuous function is not necessarily differentiable everywhere. A consequence of this is that singular points can exist. These should then be inspected separately to any critical points. Naturally, the appealing notion of a closed and finite domain means that the domain boundary (boundary points) need also to be considered separately.

## 数学代写|多变量微积分代写multivariable calculus代考|Optimization free of constraints

In relaxing the condition of compactness, either by allowing the region $R \subseteq$ $D_f$ to be unbounded or bounded but open, there is no longer any guarantee that points of finite maximum or minimum exist. For instance a function might become infinite at one or more points on the boundary of a bounded open set. Consider for example the case
$$f(x, y)=\frac{x y}{\sqrt{1-x^2-y^2}}, \quad R=D_f=\left{(x, y): x^2+y^2<1\right} .$$
The magnitude of this otherwise continuous function increases without bound as the independent variables approach the boundary of the unit disc: the function therefore attains neither an absolute maximum nor absolute minimum in $D_f$. In contrast, a continuous function on an unbounded domain may still attain a finite absolute maximum or minimum, as in the case of Mastery Check 3.7:
$$f(x, y)=\frac{4 x}{1+x^2+y^2}, \quad R=D_f=\mathbb{R}^2 .$$
The function attains both an absolute maximum and an absolute minimum despite an unbounded domain of definition.

So, how does one proceed? We need only to modify the protocol for continuous functions on compact regions. This is the right-hand side of Flowchart 3.1.

# 多变量微积分代考

## 数学代写|多变量微积分代写multivariable calculus代考|Optimization free of constraints

$$f(x, y)=\frac{4 x}{1+x^2+y^2}, \quad R=D_f=\mathbb{R}^2 .$$

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