Find the extremum of z = xy subject to x + y = 6 Watch

username2110335
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Please help with the following: Find the extremum of z = xy subject to x + y = 6

By substituting from the constraint into the objective function.
And by Lagrange's method..

Thanks
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123Master321
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(Original post by JKITFC)
Please help with the following: Find the extremum of z = xy subject to x + y = 6

By substituting from the constraint into the objective function.
And by Lagrange's method..

Thanks
I suggest you learn about Lagrange multipliers before, since this is a basic application that comes from learning
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MarcusRashford19
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Use the AM-GM inequality to find an upper bound for (xy)^0.5, hence xy
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username2110335
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(Original post by 123Master321)
I suggest you learn about Lagrange multipliers before, since this is a basic application that comes from learning
I know it... I cant do the systems of equations that it produces
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123Master321
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(Original post by JKITFC)
I know it... I cant do the systems of equations that it produces
Ok, so:


 f(x,y)=xy

g(x,y)=6 \implies g_2(x,y)=x+y-6=0
 \mathcal{L} =f(x,y) +\lambda g_2(x,y)
\mathcal{L} =xy +\lambda (x+y-6)
\frac{ \partial \mathcal{L}}{\partial x}= y+ \lambda =0
\frac{\partial \mathcal{L}}{\partial y} = x+ \lambda =0
y=-\lambda

x=-\lambda

Since x+y=6
 \lambda =-3 \implies x=y=3
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123Master321
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(Original post by MarcusRashford19)
Use the AM-GM inequality to find an upper bound for (xy)^0.5, hence xy
OP was asking for using Lagrange multipliers, otherwise you could just let y=6-x, put it into the other equation and differentiate, no need for AM-GM
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