Q 18. - Calculus

Question

Show that $f(x,y)=\sqrt{x^2+y^2}$ has a critical point $\left(0,0\right)$

Explain why f has an absolute minimum at $(0, 0)$ and why you cannot use Theorem 12.45 to show this.

Step-by-Step Solution

Answer

Determine second order derivative of the function and find the discriminate of the given function.

Step 1: Given Information

The given function is $f(x,y)=\sqrt{x^2+y^2}$

Step 2: Find the gradient and calculate the critical points

The gradient is

$\nabla f(x,y,z)=\frac{\partial f}{dx}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}$

$=\left(\frac{x}{\sqrt{x^2+y^2}}\right)\mathbf{i}+\left(\frac{y}{\sqrt{x^2+y^2}}\right)\mathbf{j}$

Function vanishes at critical points

$\nabla f(x,y)=0$

$\Rightarrow \frac{x}{\sqrt{x^2+y^2}}=0, \frac{y}{\sqrt{x^2+y^2}}=0$

The solution of above two equations are $x=0, y=0$

The critical points are $\left(0,0\right)$

Step 3: Second order derivative of function

It is given by

$\frac{{\partial }^{2}f}{\partial x^2}=\frac{y^2}{{\left(x^2+y^2\right)}^{3/2}},\frac{{\partial }^{2}f}{\partial y^2}=\frac{x^2}{{\left(x^2+y^2\right)}^{3/2}},\frac{{\partial }^{2}f}{\partial y\partial x}=-\frac{xy}{{\left(x^2+y^2\right)}^{3/2}}$

The discriminate is given by

$=\left(\frac{y^2}{{\left(x^2+y^2\right)}^{3/2}}\right)\left(\frac{x^2}{{\left(x^2+y^2\right)}^{3/2}}\right){\left(-\frac{xy}{{\left(x^2+y^2\right)}^{3/2}}\right)}^2=0$

Step 4: Conclusion

Since $H_f(x,y)=0$, no conclusion can be drawn from discriminate.

Function has terms $x^2,y^2$ , for all the points $(x,y), f(x,y)>0$, the minimum value is zero at $\left(0,0\right)$

Hence, absolute maxima is at $(0,0)$.