View Full Version : [SOLVED] Maximum value of xy
rohanprabhu
Mar25-08, 12:37 AM
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.
nicksauce
Mar25-08, 12:47 AM
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.
gamesguru
Mar25-08, 08:52 AM
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.
HallsofIvy
Mar25-08, 10:50 AM
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.
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.
rohanprabhu
Mar25-08, 09:26 PM
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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