Find min-max of function $f(x, y, z) = x + 2z$

Question

I need to find max and min (if they exist) of the following function: $f(x, y, z) = x + 2z$ On the set $C = \{(x, y, z) ∈ R^3: x + y + z = 1, x^2 + y^2 + z =7/4\}.$

I have checked that the Lagrange theorem assumptions hold, so that I can find the critical points solving the following system of equation:

$1-\lambda_1-2\lambda x=0$

$-\lambda_1-2\lambda_2 y=0$

$2-\lambda_1-\lambda_2=0$

$x+y+z=1$

$x^2+y^2+z=7/4$

From (3) we get $\lambda_1=2-\lambda_2$, that I can substitute into (2) but then I am stuck. Could you help me??

Answer 1

We have that $\nabla f \equiv (1,0,2)$. The restrictions are $g_1(x,y,z) = x+ y+z -1$ and $g_2(x,y,z) = x^2+y^2+z - 7/4$. We have hence

\begin{align} \nabla g_1(x,y,z) &= (1,1,1)\\ \nabla g_2(x,y,z) &= (2x,2y,1) \end{align}

We hence need to find solutions to $\nabla f(x,y,z) = \lambda_1\nabla g_1(x,y,z) + \lambda_2\nabla g_2(x,y,z)$ that respect the restrictions, that is

$$\left\{ \begin{align} 1 &= \lambda_1+2\lambda_2x\\ 0 &= \lambda_1 + 2\lambda_2y\\ 2 &= \lambda_1+\lambda_2\\ 1 &= x+y+z\\ 7/4 &=x^2+y^2+z \end{align} \right. $$

From the third, we get $\lambda_1 = 2-\lambda_2$. We may substitute back into the first equation to obtain $\lambda_2(2x-1) = -1$ and into the second equation to obtain $\lambda_2(2y-1) = -2$.

Any two of the first three equations imply that $\lambda_2\neq 0$, and hence $2x-1 = -1/\lambda_2$ and $2y-1 = -2/\lambda_2$. Therefore $2y-1 = 2(2x-1)$, giving $y$ in terms of $x$.

Now the fourth has $z = 1 - x - y$, and with $y$ in terms of $x$ from above we obtain $z$ in terms of $x$.

Substituting it all into the last we get a quadratic equation in $x$. Solve it, and back substitute.

Answer 2

If you subtract the first two equations, you get
$$ 2\lambda_2(x-y)=1. $$ Then $\lambda_2\ne0$, $x\ne y$. From the third equation, $$\lambda_1=2-\lambda_2=2-\frac1{2(x-y)}.$$ Now the second equation is $$ 0=\lambda_1+2\lambda_2y=2-\frac1{2(x-y)}+\frac{y}{x-y}=\frac{4x-4y-1+2y}{2(x-y)} =\frac{4x-2y-1}{2(x-y)}, $$ giving us $4x-2y=1$. Subtracting the fourth equation from the fifth, $$ x^2-x+y^2-y=\tfrac34. $$ With $y=\tfrac{4x-1}2$, this becomes $$ x^2-x+\tfrac{(4x-1)^2}4-\tfrac{4x-1}2=\tfrac34. $$ Simplifying, this becomes $x^2=x$. So $x=0$ or $x=1$. This gives us $$ y=\tfrac{4-1}2=\tfrac32,\ \ \ \ y=\tfrac{0-1}{2}=-\tfrac12. $$ And $$ z=1-x-y=1-1-\tfrac32=-\tfrac32,\ \ \ \ z=1-0-\left(-\tfrac12\right)=\tfrac32. $$ So the two critical points are $$ (1,\tfrac32,-\tfrac32)\ \ \ \ \text{ and } \ \ \ (0,-\tfrac12,\tfrac32). $$

Answer 3

The $xy$-intersection of the two given surfaces $x + y + z = 1$ and $ x^2 + y^2 + z =\frac74$ is

$$(x-\frac12)^2+(y-\frac12)^2=\frac54\tag 1$$

Rewrite $f$ in the $xy$-space as

$$f(x,y)=x+2(1-x-y) = 2-2y-x\tag 2$$

At the extreme values, its normal vector (1,2) matches that of the circle (1), i.e.

$$(x-\frac12):(y-\frac12) = 1:2$$

Substitute the resulting $y = 2x-\frac12$ into (1) to obtain the tangent points (0,-$\frac12$) and (1,$\frac32$). Plug them back into (2) to obtain the respective maximum and minimum,

$$f_{max}(0,-\frac12) = 3,\>\>\>\>\>f_{min}(1,\frac32) = -2$$