What is a way to find extreme values of a function $u(x,y,z)=xy+yz+xz$ with conditions $x+y=2, y+z=1$?
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You can turn $u(x,y,z)$ into a $1$ variable function using the substitutions $x=2-y,z=1-y$, and differentiate to find the maximum. – Meow Jun 03 '13 at 12:00
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Or note that $xy+yz+xz=(x+y)(y+z)-y^2=(2)(1)-y^2$. – Meow Jun 03 '13 at 12:03
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So, from this I get that $y=0, x=2, y=1$ and extreme value is 2? – user23709 Jun 03 '13 at 12:07
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Yes (I think you meant $z$). – Meow Jun 03 '13 at 12:08
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yes, sorry, z=1 :) – user23709 Jun 03 '13 at 12:09
2 Answers
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$$u(x,y,z)=(2-y)y+y(1-y)+(2-y)(1-y)=y(2+1-1-2)+y^2(-1-1+1)+2=2-y^2$$ Alternately, $$u(x,y,z)=(x+y)(z+y)-y^2=2-y^2$$
So the minimum is when $y=0$.
Using Lagrange multipliers,
$$g_1(x,y,z)=x+y-2=0$$ $$g_2(x,y,z)=y+z-1=0$$ $$u(x,y,z)=xy+yz+xz$$
$$\nabla u= \lambda _1 \nabla g_1+\lambda _2 \nabla g_2$$ $$(y+z,x+z,x+y)= a (1,1,0)+b (0,1,1)$$ Giving $$y+z=a$$ $$x+z=a+b$$ $$x+y=b$$ Substituting: $$x+z=x+2y+z$$ $$y=0$$
Meow
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See http://www.cs.berkeley.edu/~klein/papers/lagrange-multipliers.pdf page 5 'More dimensions'. – Meow Jun 03 '13 at 12:24
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1Thank you too, I had to learn multiple Lagrange multipliers to answer this. – Meow Jun 03 '13 at 12:33
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Notice that $2 = (x+y)(y+z) = y^2 + xy + yz + zx = y^2 + u(x,y,z)$
Hence, $u(x, y, z) \leq 2$, with equality if and only if $y=0$ (and hence $x=2, z=1$).
Calvin Lin
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