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Maximum value of xy

412
2
[SOLVED] Maximum value of xy

1. Homework Statement

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

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:

[tex]
\frac{dx}{dy} = xdy + ydx
[/tex]

also,
[tex]
dx = -dy
[/tex]

hence, for the condition of [itex]f'(x) = 0[/itex],

[tex]
xdy + ydx = 0
[/tex]

[tex]
xdy = - ydx
[/tex]

[tex]
\frac{dy}{dx} = \frac{-y}{x}
[/tex]

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

Answers and Replies

nicksauce
Science Advisor
Homework Helper
1,272
5
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.
 
85
2
This looks like a problem too easy for Lagrange multipliers, so I'll keep it simple. In case you don't know, [itex]u[/itex] is defined to be [itex]xy[/itex], so that's what we want to maximize.
[tex]x+y=0 \Rightarrow y=-x[/tex]
[tex]u=xy=-x\times x=-x^2[/tex]
Take the derivative and set to zero,
[tex]\frac{du}{dx}=0=-2x\Rightarrow x=0 \Rightarrow y=0[/tex]
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
Science Advisor
Homework Helper
41,732
893
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.
 
188
1
Perhaps it may be solved by Lagrange multiplier, you should obtain the minimum of

[tex] xy-{\lambda}(x+y) [/tex]

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

[tex] y-{\lambda}=0 [/tex]

[tex] x-{\lambda}=0 [/tex]

[tex] x+y=0 [/tex]

it seems that only a minimum at x=y=0 exists , no maximum.
 
412
2
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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