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Find the minimum value of a function $f(x,y) = x^{2} + 4y^{2} - 4xy + 3x - y + 6$ if $x+y=1$.

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up vote 3 down vote accepted

We know $y=1-x$, which we substitute into $$f(x,y) = x^{2} + 4y^{2} - 4xy + 3x - y + 6$$ to give \begin{align*} f(x,1-x) &= x^{2} + 4(1-x)^{2} - 4x(1-x) + 3x - (1-x) + 6 \\ &= 9x^2-8x+9. \end{align*} Now we can find the minimum using the usual methods for quadratics.

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So easy, huh. Thank you! –  Alexander Side Sep 22 '13 at 17:57
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If this were an intermediate Calculus problem, could use Lagrange multipliers, since x+y-1=0, if

$$F(x,y,t)=x^2+4y^2−4xy+3x−y+6+t(x+y-1)$$ $$F_x=2x-4y+3+t=0$$ $$F_y=8y-4x-1+t=0$$

Subtract to eliminate t:

$$6x-12y+4=0$$ Given $$6x+6y-6=0$$ so $$18y-10=0$$

$$y=\frac{5}{9}$$ $$x=\frac{4}{9}$$

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I think that Lagrange multiplier isn't the way to go here. It provides the exact solution, but it's more complicated, just use the much simplier substitution like Rebecca –  Stefan4024 Sep 22 '13 at 18:08
    
Both methods give exact solution, Rebecca's solves the particular problem, mine the general problem posed in the title. –  John Cosi Sep 22 '13 at 18:23
    
That's what I say. Lagrange Multiplier will evetualy provide the exact solution, but in a much more complicated way. Usually optimization problems with two variables are solved using Rebecca's way. –  Stefan4024 Sep 22 '13 at 18:50
    
@JohnCosi I won't say that your method solves the general problem, as you are making assumptions about differentiability. –  Calvin Lin Sep 22 '13 at 18:57
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