# Two Var Function

I hate min and max questions.

I have a function

G(x, y) = 1/x + 4/y + 9/(4-x-y)


I need to prove that G(x,y) has a minimum value of 9

I've partialy derived them for x and y giving

(1) 1/x^2 + 9/(4-x-y)^2 = 0
(2) 4/y^2 + 9/(4-x-y)^2 = 0


This is where I get stuck I'm not entirely sure how to continue. I need to find x = ? and y = ? but this simultaneous equation stumps me.

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Get the LCM of both terms in both equations, bring together, turn into a quadratic in x and y, eliminate... okay, where exactly are you stuck? –  Ｊ. Ｍ. Oct 30 '10 at 12:34
As it stands, your system has no solutions (a sum of two positive terms can't be zero). But that's because you made a sign error when computing the derivatives; you need to change the sign of the first term in both (1) and (2). Then note that you must have $1/x^2=4/y^2$, so that $y=\pm 2x$. From there it shouldn't be too hard, I hope! –  Hans Lundmark Oct 30 '10 at 13:03
What are possible values of x and y? e.g. G(2,3) = -7.166... < 9. –  KennyTM Oct 30 '10 at 13:16

As always, to find the maximum and minimum of a function, you have to calculate first the critical points, that is

$$\frac{\partial G(x,y)}{\partial x}=0$$ $$\frac{\partial G(x,y)}{\partial y}=0$$

In this case, the respective equations are

$$\frac{\partial G(x,y)}{\partial x}=-\frac{1}{x^{2}}+\frac{9}{(4-x-y)^{2}}$$ $$\frac{\partial G(x,y)}{\partial y}=-\frac{4}{y^{2}}+\frac{9}{(4-x-y)^{2}}$$

Simplifying you should obtain $x=1-\displaystyle \frac{y}{4}$ and $\displaystyle \frac{5y}{2}=4-x$ which lead to the solution $(x,y)=\bigg\{\displaystyle \frac{2}{3},\displaystyle \frac{4}{3}\bigg\}$

Then you substitute those values in your original function $G(\frac{2}{3},\frac{4}{3})=9$.

As I said, such a point is just a critical point (which means, it could be a local minimum, a local maximum or a saddle point). In order to figure out, we have to use the Second partial derivative test.

You can see the details in the Wikipedia article. As you already know what you want to prove (that $(\frac{2}{3},\frac{4}{3})$ is a local minimum), we expect that $M(\frac{2}{3},\frac{4}{3})>0$ and $f_{xx}(\frac{2}{3},\frac{4}{3})>0$.

I think I'll leave it there. You can do the rest of the calculations.

By the way, the aforementioned procedure works to obtain only a local minimum, maximum or saddle point, not global extrema.

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You cannot prove that the minimum is 9, because the function certainly takes on lower values: for example $G=0$ along a the parabola $(x,y)=(-2t+6t^2,8t+12t^2)$, and clearly $G\rightarrow 0$ as $x,y \rightarrow\infty$. What you have discovered is that there is only one LOCAL minimum, and it is at $(2/3, 4/3)$.

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