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I have some problem about physics, which requires me to find the lowest value of $f(x,y)=x^2 + 3y^2$ while $x+3y=12$.

I have used the lagrange multiplier but I'm not sure whether it gave me the minima or the maxima, so how can I find the lowest value of $f(x,y)$ with this restraint ?

$$f(x,y)=x^2 + 3y^2$$ and $$g(x,y)=x+3y$$

Since $$\frac{\partial f}{\partial x} = 2x = \lambda \frac{\partial g}{\partial x} = \lambda $$ $$\frac{\partial f}{\partial y} = 6y = \lambda \frac{\partial g}{\partial x} = \lambda 3$$

we get $x=y$, but I don't know whether it is a minima or the maxima, so how can I check that ?

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  • $\begingroup$ Could you show your work using Lagrange multipliers ? It works. $\endgroup$ – Claude Leibovici Dec 4 '16 at 10:06
  • $\begingroup$ @ClaudeLeibovici sure. $\endgroup$ – onurcanbektas Dec 4 '16 at 10:06
  • $\begingroup$ Since you solve a convex optimization problem, you found a minimum. Use $x+3y=12$ with $x=y$ to find the value of $x$ and $y$. $\endgroup$ – LinAlg Dec 4 '16 at 12:06
  • $\begingroup$ @LinAlg How do you know that it is a minimum ? Lagrange multiplier gives you a extremum. $\endgroup$ – onurcanbektas Dec 4 '16 at 12:39
  • $\begingroup$ Not for convex problems. See the (free) book by Boyd and Vandenberghe. $\endgroup$ – LinAlg Dec 4 '16 at 14:05
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with $$y=4-\frac{1}{3}x$$ you will get $$f(x,4-1/3x)=x^2+3(4-1/3x)^2$$ a problem in one variable.

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Use Cauchy-Schwarz inequality: $144=12^2 = (x+3y)^2 = (1\cdot x + \sqrt{3}\cdot \sqrt{3}y)^2 \le (1^2+\sqrt{3}^2)(x^2+3y^2)=4(x^2+3y^2)\implies x^2+3y^2 \ge \dfrac{144}{4}=36 $, and this is the minimum value.

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  • $\begingroup$ CS inequality ? $\endgroup$ – onurcanbektas Dec 4 '16 at 10:10
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If $f(x,y)=x^2+3y^2$ is to be minimzed subject to $x+3y=12$ then this is equivalent to minimizing $$g(y)=(12-3y)^2+3y^2\tag{1}$$ subject to $y$.

The first order condition for a solution $y^*$ to $(1)$ is that $$g'(y^*)=0;\tag{2}$$ and the second order condition is that $$g''(y^*)\leq 0\tag{3}.$$

In general, both $(2)$ and $(3)$ must be satisfied for a local minimum. In this case, however, $g(y)$ is a quadratic function and you may write $g(y)$ as $a(y-b)^2+c$ for constants $a$, $b$ and $c$, and then notice that the minimum is attained at $y=b$.

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Just put in $x=12-3y$ into the first equation. it is easy to find the lowest value of a quadratic function of $y$.

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