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The prompt is to find the extreme values of an implicit function $z(x, y)$

The functions are

  • $$ x^2 + y^2 + z^2 -3z = 0$$

  • $$x^2 + y^2 +z^2 -2x -2z +2 =0$$

Solving functions with just 2 variables, I know that we are supposed to start by finding partial derivatives of each variable and equate them to find 2 values that helps find the critical points, following the same lines, for the first part I found partial derivatives as $$f_x = 2x$$ $$f_y = 2y$$ $$f_z = 3z - 3$$ $$f_{xx} = 2$$ $$f_{yy} = 2$$ $$f_{zz} = 3$$ $$f_{xyz} = 0$$ But I'm not sure how to proceed finding the extreme values

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3 Answers 3

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first of all, your surfaces is a sphere, and you should know how to solve this problem via geometry.

But since you are learning calculus, use your knowledge of geometry to guide you / check your answers.

$\frac {\partial }{\partial x}(x^2 + y^2 + z^2 - 3z = 0)\\ 2x + (2z - 3) \frac {\partial z}{\partial x} = 0)\\ \frac {\partial z}{\partial x} = \frac {x}{z-\frac 32} = 0$

at $x = 0$

similarly $\frac {\partial z}{\partial y} = \frac {y}{z-\frac 32} = 0$ at $y = 0$

you have critical points $(0,0,3), (0,0,0)$

now find $\frac {\partial^2 z}{\partial x^2},\frac {\partial^2 z}{\partial y^2}, \frac {\partial^2 z}{\partial x\partial y}$

at the critical points identified

I get $-\frac {2}{3}, -\frac {2}{3}, 0$ and $\frac {2}{3}, \frac {2}{3}, 0$

finally:

$4(\frac {\partial^2 z}{\partial x^2})(\frac {\partial^2 z}{\partial x^2})>(\frac {\partial^2 z}{\partial x\partial y})^2$

For both points, suggesting neither is a saddle.

min at $(0,0,0)$ max at $(0,0,3)$

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  • $\begingroup$ How were the critical points $(0, 0, 3)$ and $(0, 0, 0)$ were identified? $\endgroup$ Jun 27, 2017 at 2:43
  • $\begingroup$ we found the values of $x, y$ such that $\frac {\partial z}{\partial x} = \frac {\partial z}{\partial y} = 0 $ and solved for $z.$ $\endgroup$
    – Doug M
    Jun 27, 2017 at 2:50
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From the first equation, you could get that $$x^{2} + y^{2} = 3z - z^{2}$$ and then plug in the second equation, and you will get that $$z - 2x + 2 =0$$

Now , you could express $z$ by $x$.

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

Taking the derivative could give you some potential points to obtain extreme value. Be cautious, check the Hessian matrix.

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    $\begingroup$ I think these are two different questions, not a system of equations. Am I right, @Ritwick? $\endgroup$ Jun 26, 2017 at 21:39
  • $\begingroup$ The two equations are 2 separate parts of the question $\endgroup$ Jun 26, 2017 at 21:39
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Without calculus:

$1)$ The roots of the quadratic equation $z^2-3z+(x^2+y^2)=0:$ $$z=\frac{3\pm\sqrt{9-4x^2-4y^2}}{2}.$$ $$0\le 9-4x^2-4y^2\le 3 \Rightarrow 0\le z\le 3.$$

$2)$ Rearrange: $$(x-1)^2+y^2+(z-1)^2=0.$$ It is a sphere with center at $(1; 0; 1)$ and radius $0$.

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