Max and min value of $f(x,y)$

Question

$$f(x,y) = 2\cdot x\cdot y$$ $$x^2+y^2 \leq 4$$ I have no idea about this question.There is a region.How can I solve?

Answer 1

At first make two cases: $$x^2+y^2 <4$$ this one is open hence you need the gradient to be zero there.

The second case is the boundary so $$x^2+y^2=4$$ Here take langrange multipliers

Answer 2

Here is a tedious way to solve it:

First, the set $C = \{(x,y)| x^2+y^2 \le 4 \}$ is compact, so you know the extrema are attained. The objective is smooth.

If the extrema is attained in $C^\circ$, then the gradient must be zero, which results in $(x,y) = (0,0)$. However, since $(1,1)$ and $(1,-1)$ result in values greater than or less than, we see that $(0,0)$ is neither a minimum nor maximum.

Hence the extrema are attained on $\partial C = \{(x,y)| x^2+y^2 = 4 \}$. Since the gradient of $g(x,y) = x^2+y^2-4$ is non-zero on $\partial C$, we can use Lagrange multipliers, which results in $$2\binom{y}{x}+ \lambda 2\binom{x}{y} = 0$$ This gives $y = (-\lambda)^2 y$ and $x = (-\lambda)^2 x$, from which it follows that $\lambda \in \{\pm 1 \}$, and hence $x = \pm y$. This restricts the possibilities to $x,y \in \{\pm \sqrt{2} \}$, from which we obtain the minimum of $-4$ and a maximum of $+4$.

Here is a slightly less tedious way:

Let $f(r,\theta) = r(\cos \theta, \sin \theta)$. Then $C = \{ f(r,\theta) | r \in [0,2], \theta \in \mathbb{R} \}$. Hence the problem is equivalent to extremizing $2(r \cos \theta )(r \sin \theta)$, subject to $(r,\theta) \in [0,2]\times \mathbb{R}$. Since $2(r \cos \theta )(r \sin \theta) = r^2 \sin ( 2 \theta)$, and $\sin ( 2 \theta) \in [-1,1]$, it follows that the extreme values are $\pm 4$.

Answer 3

$(x-y)^2 \ge 0$

$x^2+y^2-2|x||y| \ge0$

$x^2+y^2 \le 4$

Maximum value of $x^2+y^2=4$

Therefore, $4-2xy \ge 0 \implies4 \ge2|x||y|$

Negating the same inequality yeilds:

$-4 \le -2|x||y|$

$-4 \le 2|(-x)||y| \implies -4 \le 2|x||y|$

Answer 4

Since $(|x|-|y|)^2\geq 0$, we have $$ 2|xy|\leq x^2+y^2\leq 4. $$ Hence $$ -4\leq 2xy\leq 4. $$ If $x=y=\sqrt{2}$ we have $x^2+y^2=4\leq 4$ and $2xy=4$.

If $x=\sqrt{2}, y=-\sqrt{2}$ we have $x^2+y^2=4\leq 4$ and $2xy=-4$.

Therefore, the minimum value of $2xy$ is $-4$ and the maximum value of $2xy$ is $4$.

Answer 5

Hint :Use A.M - G.M inequality for $|x|,|y|$.

$\Rightarrow x^2+y^2\ge2|xy|$

$\Rightarrow 4\ge x^2+y^2\ge 2|xy|$

$\Rightarrow-4\le2xy\le 4$

Equality holds when $|x|=|y|=\sqrt{2}$