Use Lagrange Multipliers to find max and min of $x+2y$ subject to $x+y+z=1$ and $y^2+z^2=4$

Question

Using Lagrange Multipliers, determine the maximum and minimum of the function $f(x,y,z) = x + 2y$ subject to the constraints $x + y + z = 1$ and $y^2 + z^2 = 4$: Justify that the points you have found give the maximum and minimum of $f$.

So, $$ \nabla f = (λ_1)\nabla g_1 + (λ_2)\nabla g_2 $$

I get to this point $$ (1,2,0) = λ_1(1,1,1) + λ_2(0,2y,2z) $$ Where do I go from here to find the critical points ect.

Answer 1

We can find the extreme values using one constraint only: $z = 1 - x - y$, so $y^2 + (1-x-y)^2 = 4$.

Thus: $g(x,y) = y^2 + 1 + x^2 + y^2 - 2x - 2y + 2xy - 4 = 0$.

So: $\nabla f = \lambda\nabla g$ gives:

$(1,2) = \lambda(2x-2+2y,4y-2+2x)$. Thus:

$\dfrac{1}{2x-2+2y} = \dfrac{2}{4y-2+2x}$. Hence:

$4y-2+2x = 4x - 4 + 4y$. So $x = 1$, and $y^2 + y^2 = 4$. So: $y^2 = 2$, and $y = \pm \sqrt{2}$. Thus we easily see that extrema are:

$f_{min} = f(1,-\sqrt{2}) = 1 - 2\sqrt{2}$, and $f_{max} = f(1,\sqrt{2}) = 1 + 2\sqrt{2}$.

Answer 2

I get to this point $$ (1,2,0) = λ_1(1,1,1) + λ_2(0,2y,2z) $$

Componentwise we have $1=\lambda_1$, $2=\lambda_1+2\lambda_2y$, $0=\lambda_1+2\lambda_2z$. Hence, $\lambda_1=1$, $y=\frac1{2\lambda_2}$, $z=-\frac1{2\lambda_2}$, e.i., $z=-y$.

From the constraint $y^2+z^2=4$ and $z=-y$, we have $y=\pm\sqrt2$; from the constraint $x+y+z=1$ and $z=-y$, we have $x=1.$

Conclusion: $x+2y=1\pm\sqrt2$ are extrema. It is not so obvious but $1+\sqrt2$ is max. and $1-\sqrt2$ is min. One may need a second derivative test, Hessian matrix etc.

Answer 3

Here is another solution with the completing square technique.

Let's denote $L = x+ 2y$, then $x = L - 2y$. From the first condition, we have $z = 1-L + y$ and put it to the second condition:

$$\begin{align} 4 &= y^2 + z^2 \\ &= y^2 +(1-L+y)^2\\ &=2\left(y^2 + (1-L)y + (1-L)^2 \right)\\ 4&=2\left(y + \frac{1-L)}{2}\right)^2 + \frac{(1-L)^2}{2} \ge \frac{(1-L)^2}{2} \tag{1} \end{align}$$ From $(1)$, we deduce then $$(1-L)^2\le 8 \iff 1-\sqrt 2 \le L \le 1+\sqrt 2$$

The equality for the min (resp. max) occurs if and only if $(x,y,z) = (1, -\sqrt 2, 0)$ (resp. $(x,y,z) = (1, \sqrt 2, 0)$ ).