# Minimize the function $f(y_1,y_2)=3 y_1^2+8y_2^2$ [closed]

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?

• What went wrong when using Lagrange multiplies? Wolfram Alpha can solve it just fine (i.sstatic.net/JY5EW.png, rename $y_1 = x, y_2 =y$). Commented Jun 20, 2016 at 2:43
• Lagrange multipliers do work just fine here. Commented Jun 20, 2016 at 2:47
• Could explicit with Lagrange multipliers Commented Jun 20, 2016 at 2:48
• @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. Commented Jun 20, 2016 at 2:51

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 $y_1=\pm 1$.
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 $(0,1)$.
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.
• Take the derivative wrt $y_2$ , set it equal to zero I guess Commented Jun 20, 2016 at 2:50
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 $$3\le f \le 8$$