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I would like to minimize $f(y_1,y_2)=3 y_1^2+8y_2^2$ with the constraints $g(y_1,y_2)=y_1^2+y_2^2=1$. I thought I could use the Lagrange multipliers, but it is not work. Is there anyone could show me how to find it?

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    $\begingroup$ What went wrong when using Lagrange multiplies? Wolfram Alpha can solve it just fine (i.sstatic.net/JY5EW.png, rename $y_1 = x, y_2 =y$). $\endgroup$ Commented Jun 20, 2016 at 2:43
  • $\begingroup$ Lagrange multipliers do work just fine here. $\endgroup$
    – Ethan Hunt
    Commented Jun 20, 2016 at 2:47
  • $\begingroup$ Could explicit with Lagrange multipliers $\endgroup$
    – Sharpie
    Commented Jun 20, 2016 at 2:48
  • $\begingroup$ @Sharpie If you are interested in getting help concerning the method, it would be a good idea to show at least how you set up your "Lagrange equations". It would give us an idea of why you're not getting this to work. $\endgroup$ Commented Jun 20, 2016 at 2:51

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Personally, I don't know what Lagrange multipliers are, so this could be wrong, but: $$f(y_1,y_2)=3y_1^2+8y_2^2=3y_1^2+3y_2^2+5y_2^2=3+5y_2^2$$ Thus, the minimum is $3$ at $y_2=0$ and $y_1=\pm 1$.

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    $\begingroup$ Good! Before worrying about what tool one should use, one looks at the problem. $\endgroup$ Commented Jun 20, 2016 at 2:50
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Lagrange multipliers work well. Let \begin{align} f(x,y)&=3x^2+8y^2\\ g(x,y)&=x^2+y^2=1, \end{align} then there is $\lambda\in \mathbb{R}$ such that $\nabla f = \lambda \nabla g$. That is, \begin{align} 6x &= 2x\lambda\\ 16y &= 2y \lambda \end{align} and so $2x(3-\lambda)=0$ and $2y(8-\lambda)=0$. If $x=0$, then $y=\pm 1$, so $\lambda$ must be $8$. If $y\ne 0$, then $x=\pm 1$, so $\lambda$ must be $3$. If $xy\ne 0$, then one of $3-\lambda$ and $8-\lambda$ can't be zero, a contradiction. Thus we can know that minimum of $f$ with the constraint $g$ is $3$ at $(-1,0)$ and $(1,0)$, and maximum $8$ at $(0,-1)$ and $(0,1)$.

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from the constraint you have that $3y_1 + 3y_2 = 3$. now you write the function as $3y_1+3y_2+5y_2 = f(y_1,y_2)$. Hence you get $f(y_2) = 3 + 5y_2$. Here you use simple max/min technique from one variable calculus.

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  • $\begingroup$ Take the derivative wrt $y_2$ , set it equal to zero I guess $\endgroup$ Commented Jun 20, 2016 at 2:50
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Put $y_1=sin(t)$ and $y_2=cos(t)$. These substitions satisfy your constraints. So put these values in $f$, then easily interpret the maximum and minimum value of the function. $f$ will lie between $$3\le f \le 8$$

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