Chapter 12: Q 18. (page 975)

Show that $f (x, y) = \sqrt{x^{2} + y^{2}}$ has a critical point $(0, 0)$
Explain why f has an absolute minimum at $(0, 0)$ and why you cannot use Theorem 12.45 to show this.

Short Answer

Expert verified

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

Step by step solution

Given Information

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

Find the gradient and calculate the critical points

The gradient is

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

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

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 $(0, 0)$

Second order derivative of function

It is given by

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

The discriminate is given by

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

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 $(0, 0)$

Hence, absolute maxima is at $(0, 0)$ .

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Show that $f (x, y) = \sqrt{x^{2} + y^{2}}$ has a critical point $(0, 0)$
Explain why f has an absolute minimum at $(0, 0)$ and why you cannot use Theorem 12.45 to show this.

Short Answer

Step by step solution

Given Information

Find the gradient and calculate the critical points

Second order derivative of function

Conclusion

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Show that f(x,y)=x2+y2has a critical point 0,0Explain why f has an absolute minimum at (0,0)and why you cannot use Theorem 12.45 to show this.

Short Answer

Step by step solution

Given Information

Find the gradient and calculate the critical points

Second order derivative of function

Conclusion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Probability and Statistics

Theoretical and Mathematical Physics

Geometry

Mechanics Maths

Statistics

Study anywhere. Anytime. Across all devices.

Show that $f (x, y) = \sqrt{x^{2} + y^{2}}$ has a critical point $(0, 0)$
Explain why f has an absolute minimum at $(0, 0)$ and why you cannot use Theorem 12.45 to show this.