# Relative max and Min problem

#### mike21

Seem to be having trouble with this one. Can someone please walk me trough/solve this. Find and identify all rel max and min of F(x,y)=xy+2/x+4/y.

My partial deriv. in respect to x is y-2x^-2=0
in respect to y is x-4y^-2=0, but im not even sure if thats right. Then I think you have to solve for x and y, and then at some point take the second partial deriv, but I just can't get it.

#### chisigma

MHF Hall of Honor
Let $$\displaystyle f(x,y) = x y + \frac{2}{x} + \frac{4}{y}$$ and $$\displaystyle \beta$$ an arbitrary real number, we can ever find two values $$\displaystyle x_{0}$$ and $$\displaystyle y_{0}$$ so that is $$\displaystyle f(x_{0}, y_{0}) > \beta$$ or $$\displaystyle f(x_{0}, y_{0}) < \beta$$ and that means that $$\displaystyle f(x,y)$$ is umbounded and it hasn't neither maxima nor minima...

Kind regards

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### HallsofIvy

MHF Helper
Apparently chisigma missed the "rel" as well as "Relative" in the title of the thread. The fact that the function is unbounded means it has no absolute (global) max or min but does not mean it cannot have relative (local) max and min.

Mike21, yes, the partial derivatives of xy+ 2/x+ 4/y are $$\displaystyle y- 2/x^2$$ and $$\displaystyle x- 4/y^2$$ and those are set to 0. That means that we have $$\displaystyle y= 2/x^2$$, so $$\displaystyle x^2y= 2$$ and $$\displaystyle x= 4/y^2$$, so $$\displaystyle xy^2= 4$$. Dividing the second equation by the first (if x or y equals 0, the original function is not defined) we have $$\displaystyle \frac{y}{x}= 2$$ or $$\displaystyle y= 2x$$. Put that back into $$\displaystyle x^2y= 2$$ to find the critical point and check to see if it is a relative max or min.

• chisigma