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I am trying to calculate the local maximum and minimum of $f(x, y) = x$$3$$y$$2$$(2 − x − y)$, however I seem to keep running into issues.

My steps were to find the partial derivative for $f$$x$$(x,y)$ and $f$$y$$(x,y)$, then find the values for y and substitute them back to find x. After that I found $D=f$$xx$$(x,y)$$f$$yy$$(x,y)$-[$f$$xy$$(x,y)$ ]2 to determine the type of stationary point.

I think most of my problems stem from incorrectly identifying the stationary points to begin with, any help would be appreciated.

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If you determin gradient of $f(x,y)$ you will get $$[6x^2y^2-4x^3y^2-3x^2y^3,4x^3y-2x^4y-3x^3y^2]$$ $$[x^2y^2(6-4x-3y),x^3y(4-2x-3y)]$$ Finding the stationary points will give you $(x,y)=(0,0)$ and $(x,y)=(1,\frac{2}{3})$

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  • $\begingroup$ I found the (0,0) roots by simply by concluding that outer y must equal zero, but I can't get the inside to give me the solution y = 1 $\endgroup$ – Lauren Sep 9 '15 at 2:24
  • $\begingroup$ What exactly do you mean? You just have to solve the system of linear equations $6-4x-3y=0$ and $4-2x-3y=0$. $\endgroup$ – MrYouMath Sep 9 '15 at 7:40

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