Algebraically Sound? Multivariate Extrema step equating partials

[boldimage]

Hi guys,

I just had a quick question with regards to a step I am up to in a multivariate question.

It asks me to find the local extrema for:

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

I know I have to equate the partials and use the second derivative test but I get to:

f_x = 3x^2 - 8y

f_y = 3y^2 - 8x

I set both to zero but don't know how to solve them simultaneously to get the point.

I tried x(3x-8y) = x(3y^2-8) and then thought I could maybe cancel the x's away and then factor out the y's and do the same but I get x=y in the end?

Any help would be greatly appreciated!

Thanks, M.

 

[tkhunny]

This is not good. How do you get to multi-variable calculus and forget algebra?!

1) Take a quick glance at it and observe that x = y = 0 is a solution.

2) Have you considered substitution? \(\displaystyle y = \frac{3}{8}x^{2}\) so that \(\displaystyle 8x = 3\left(\frac{3}{8}x^{2}\right)^{2}\)

Please get your algebra up tp speed before you have to bail on the whole class. How's your trigonometry?

 

[soroban]

Hello, boldimage!

Here's another approach . . .

\(\displaystyle \text{Find the local extrema for: }\:f(x,y) \:=\: x^3 +y^3 -8xy\)

I know I have to equate the partials and use the second derivative test,
but I get to: .\(\displaystyle \begin{Bmatrix}f_x \:=\: 3x^2 - 8y \\ f_y \:=\: 3y^2 - 8x\end{Bmatrix}\)

I set both to zero, but don't know how to solve them simultaneously to get the points.

We have: .\(\displaystyle \begin{Bmatrix}3x^2 - 8y \:=\:0 & [1] \\ 3y^2 - 8x \:=\:0 & [2] \end{Bmatrix}\)

Then: .\(\displaystyle 3x^2 - 8y \:=\:3y^2 - 8x \quad\Rightarrow\quad 3x^2 - 3y^2 + 8x - 8y \:=\:0\)

. . \(\displaystyle 3(x^2-y^2) + 8(x-y) \:=\:0 \quad\Rightarrow\quad 3(x-y)(x+y) + 8(x-y) \:=\:0\)

. . \(\displaystyle (x-y)\big[3(x+y) + 8\big] \:=\:0 \)

So we have: .\(\displaystyle \begin{Bmatrix}x - y \:=\:0 & \Rightarrow & x \:=\:y & [3] \\ 3(x + y) + 8 \:=\:0 & \Rightarrow &y \:=\: \text{-}x-\frac{8}{3} & [4]\end{Bmatrix}\)


Substitute [3] into [1]: .\(\displaystyle 3x^2 - 8x \:=\:0 \quad\Rightarrow\quad x(3x - 8) \:=\:0\)

. . And we have: .\(\displaystyle \begin{pmatrix}x\:=\:0 \\ y\:=\:0\end{pmatrix}\:\text{ and }\:\begin{pmatrix}x \:=\:\frac{8}{3} \\ y \:=\:\frac{8}{3}\end{pmatrix}\)


Substitute [4] into [1]: .\(\displaystyle 3x^2 - 8\left(\text{-}x-\frac{8}{3}\right) \:=\:0 \quad\Rightarrow\quad 3x^2 + 8x + \frac{64}{3}\:=\:0 \)

. . But .\(\displaystyle 9x^2 + 24x + 64 \:=\:0\) .has no real roots.


Therefore, the extrema are: .\(\displaystyle (0,0,0)\:\text{ and }\:\left(\frac{8}{3},\:\frac{8}{3},\:-\frac{512}{27}\right)\)