7. does $f(x,y):=sqrt{x^{2}+y^{2}}$ have any critical points? where is the absolute minimum of this function…

Explanation:

Step1: Find the first - order partial derivatives

Let $f(x,y)=\sqrt{x^{2}+y^{2}}$. The first - order partial derivatives are: $\frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}}$ and $\frac{\partial f}{\partial y}=\frac{y}{\sqrt{x^{2}+y^{2}}}$. Set $\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$. $\frac{x}{\sqrt{x^{2}+y^{2}}}=0$ implies $x = 0$, and $\frac{y}{\sqrt{x^{2}+y^{2}}}=0$ implies $y = 0$. So, the critical point is $(0,0)$.

Step2: Consider the boundary

The boundary is given by $x^{2}+y^{2}=10$. Let $x=\sqrt{10}\cos t$ and $y = \sqrt{10}\sin t$. Then $f(x,y)=\sqrt{x^{2}+y^{2}}=\sqrt{10\cos^{2}t + 10\sin^{2}t}=\sqrt{10}$. At the critical point $(0,0)$, $f(0,0)=\sqrt{0^{2}+0^{2}}=0$.

Step3: Determine the absolute minimum

Since $0\leq\sqrt{x^{2}+y^{2}}\leq\sqrt{10}$ for $x^{2}+y^{2}\leq10$, the absolute minimum value of $f(x,y)$ is $0$ which occurs at the point $(0,0)$.

Answer:

The critical point is $(0,0)$ and the absolute minimum value of the function $f(x,y)$ in the region $x^{2}+y^{2}\leq10$ is $0$ at the point $(0,0)$.