# Maximum value of xy

1. Mar 25, 2008

### rohanprabhu

[SOLVED] Maximum value of xy

1. The problem statement, all variables and given/known data

Q] Given that $x \in [1, 2]$ and $y \in [-1, 1]$ and $x + y = 0$, find the maximum value of $xy$

3. The attempt at a solution
I have no idea at all. Does this have something to do with the maxima/minima. In that case, i can get that:

$$\frac{dx}{dy} = xdy + ydx$$

also,
$$dx = -dy$$

hence, for the condition of $f'(x) = 0$,

$$xdy + ydx = 0$$

$$xdy = - ydx$$

$$\frac{dy}{dx} = \frac{-y}{x}$$

i don't even know what i'm doing till now.

2. Mar 25, 2008

### nicksauce

If f = xy, and x + y = 0, then f = -x^2. I think it should be fairly straight forward to find the maximum value of this function. Of course if this is not in the region for which the function is defined, then you just need to check at the boundaries.

3. Mar 25, 2008

### gamesguru

This looks like a problem too easy for Lagrange multipliers, so I'll keep it simple. In case you don't know, $u$ is defined to be $xy$, so that's what we want to maximize.
$$x+y=0 \Rightarrow y=-x$$
$$u=xy=-x\times x=-x^2$$
Take the derivative and set to zero,
$$\frac{du}{dx}=0=-2x\Rightarrow x=0 \Rightarrow y=0$$
This makes sense because it's going to be the product of a negative number and its absolute value. So the largest is going to be at zero.

4. Mar 25, 2008

### HallsofIvy

Staff Emeritus
In fact you don't need to differentiate at all. Once you realize that u(x)= -x2, it is clear that u is negative for all x except x= 0.

5. Mar 25, 2008

### mhill

Perhaps it may be solved by Lagrange multiplier, you should obtain the minimum of

$$xy-{\lambda}(x+y)$$

differenentiating respect to x , y and lambda we get the equations

$$y-{\lambda}=0$$

$$x-{\lambda}=0$$

$$x+y=0$$

it seems that only a minimum at x=y=0 exists , no maximum.

6. Mar 25, 2008

### rohanprabhu

thanks to everybody.. i got it now. I really feel stupid about this problem. I have no idea about Lagrange multiplier, but calculating f(x) is something i should've done... thanks to everyone again.

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