Critical points question

[mooshupork34]

The following problem is an example in my practice book, but I'm still not clear about the way my book explained how to find the solution. If anyone could explain how the problem is done step by step, it would be greatly appreciated!

Find and classify all critical points of the following function:

f(x,y) = x^3 - 6xy + 8y^3

 

[galactus]

Start by finding the partial derivatives of f(x,y).

\(\displaystyle \L\\f_{x}=3x^{2}-6y=0\)........[1]

\(\displaystyle \L\\f_{y}=24y^{2}-6x=0\)........[2]

Same as with one variable, a function of two variables has critical points in the interior of the domain of f where
\(\displaystyle \L\\f_{x}(x,y)=0 \;\ and \;\ f_{y}(x,y)=0\)

If you solve [1] for y, you get \(\displaystyle \L\\y=\frac{x^{2}}{2}\)

Sub this into [2] and we find that \(\displaystyle \L\\x=1 \;\ or \;\ 0\)

Resubbing gives \(\displaystyle \L\\y=\frac{1}{2} \;\ or \;\ 0\)

There are extrema at \(\displaystyle \L\\(1,\frac{1}{2}), \;\ (0,0)\)

You can use the second partials to test the extrema.

 

[soroban]

Hello, mooshupork34!

I made a terrible blunder in this post . . . I'll edit it now.
[Thanks for the heads-up, Cody!]

Find and classify all critical points of the following function:
. . . \(\displaystyle f(x,y) \:=\: x^3\,-\,6xy\,+\,8y^3\)

Find the two partial derivatives and equate to zero.

. . \(\displaystyle \begin{array}{cccccc}f_{x} & \:=\: & 3x^2\,-\,6y & \:=\: & 0 &\; [1] \\
f_y & = & -6x\,+\,24y^2 & = & 0 & \;[2] \end{array}\)

From [2], we have: \(\displaystyle \:x \:=\:4y^2\;\;[3]\)

Substitute into [1]: \(\displaystyle \:3(4y^2)^2\,-\,6y\:=\:0\;\;\Rightarrow\;\;48y^4\,-\,6y\:=\:0\)

. . and we have: \(\displaystyle \:6y(8y^3\,-\,1) \:=\:0\;\;\Rightarrow\;\;y \:=\:0,\,\frac{1}{2}\)

Substitute into [3] and we get: \(\displaystyle \:x\:=\:0,\,1\)

. . There are two critical points: \(\displaystyle \L\0,\:0),\;\left(1,\:\frac{1}{2}\right)\)


Second Partial Test: \(\displaystyle \\;=\;\left(f_{xx}\right)\left(f_{yy}\right)\,-\,\left(f_{xy}\right)^2\)

We have: \(\displaystyle \:\begin{array}{ccc}f_{xx} & = & 6x \\ f_{yy} & = & 48y \\ f_{xy} & = & -6\end{array}\)

. . Hence: \(\displaystyle \ \;=\;(6x)(48y)\,-\,(-6)^2 \;=\;288xy\,-\,36 \;=\;36(8xy\,-\,1)\)


At \(\displaystyle (0,\,0):\;D \:=\:36(8\cdot0\,-\,1)\:=\:-36\) . . . negative: concave down, \(\displaystyle \cap\)

. . Therefore, a relative maximum at \(\displaystyle (0,\,0,\,0)\)


At \(\displaystyle \left(1,\,\frac{1}{2}\right):\;D\:=\:36\left(4\cdot\frac{1}{2}\,-\,1\right) \:=\:+108\) . . . positive: concave up, \(\displaystyle \cup\)

. . Therefore, a relative minimum at \(\displaystyle \left(1,\,\frac{1}{2},\,-1\right)\)


I hope it's right this time . . .
.