Constraints On Cylinders And Planes

This is a solution without using calculus, but only inequalities:

$\displaystyle\frac{x+y}{2}\leq\sqrt{\frac{x^2+y^2}{2}}=\frac{\sqrt{2}}{2}$ because of $AM-QM$ and this implies $x+2y+z=x+y+1\leq\sqrt{2}+1=M$ . We get the equality with $x=y=1-z=\frac{1}{\sqrt{2}}$

$\displaystyle-\frac{-x-y}{2}\geq-\sqrt{\frac{(-x)^2+(-y)^2}{2}}=-\frac{1}{\sqrt{2}}$ always because of $AM-GM$ , so $x+2y+z\geq1-\sqrt{2}=m$ . The values for the equality are $x=y=1-z=-\frac{1}{\sqrt{2}}$ .

Then the solution is $\boxed{M+m=2}$

Another solution, which, admittedly, works because of the specific objective and constraints of the problem.

Since $x^2+y^2=1$ , let $x=\cos \theta, \ y=\sin \theta$ . Note that because of the given constraint $y+z=1$ , the objective function becomes just $x+y+1$ . Since $x+y=\cos\theta+\sin \theta=\sqrt{2}\sin (\theta+\pi/4)\in [-\sqrt{2},\sqrt{2}]$ ,we have $M=\sqrt{2}+1,\ m=1-\sqrt{2}$ . Thus $M+m=\boxed{2}$ .

For the target function $f(x,y,z)=x+2y+z$ first we define a function by $\displaystyle g(x,y,z)_{\lambda_1,\lambda_2} = x+2y+z-\lambda_1(x^2+y^2-1)+\lambda_2(y+z-1)$

On obtaining the first partial derivatives we have : $\displaystyle \begin{cases} 1-2\lambda_1 x=0 \\ 2-2\lambda_1 y-\lambda_2=0 \\ 1-\lambda_2=0 \\ x^2+y^2=1 \\ y+z=1 \end{cases}$ , Solving upon which you get $\displaystyle x=y=\pm \frac{1}{\sqrt{2}}$ & $z=1-y$

For $x,y>0$ we get the maximum as $f(x,y,z)\le 1+\sqrt{2}$ and for $x,y<0$ we have $f(x,y,z)\ge 1-\sqrt{2}$ and thus $\displaystyle 1-\sqrt{2}\le f(x,y,z) \le 1+\sqrt{2}$ which makes $\boxed{\text{M+m=2}}$

$\text{Note : }$ The verification of the extremums are justified since the matrix(specifically Hessian Matrix ) $\displaystyle \begin{pmatrix} -2\lambda_1 & 0 & 0 \\ 0 & -2\lambda_1 & 0 \\ 0&0&0 \end{pmatrix}$ , the sign of whose entries depends upon chosen $x,y$ . For $x,y>0$ we have $\lambda_1>0$ and thus the matrix has negative eigenvalues and so it's negative definite which implies existence of a maximum.

Similarly for $x,y<0$ , $\lambda_1<0$ and so the entries of the matrix are positive and the eigenvalues are all positive which makes it a positive definite in nature implying the existence of a minimum. So the function given is bounded in the circle $x^2+y^2=1$ & the plane $y+z=1$ .

Lastly thanks for dedicating a problem to me :) .

When you add $2xy$ to $x^2+y^2=1$ , you get $x^2+2xy+y^2=2xy+1$ . You can therefore square root both sides, getting: $|x+y|=\sqrt{2xy+1}$ . You can remove the absolute by doing: $x+y=\pm\sqrt{2xy+1}$ . If you add this to the other equation, you get: $x+2y+z=\pm\sqrt{2xy+1}+1$ . You would want to maximize the square root for both maximum and minimum, as you can use the $\pm$ to get the value that is desired. Therefore, the maximum and minimum will have the same value for $\sqrt{2xy+1}$ , but you just use different signs. Therefore, without even finding the values, you get that the maximum and minimum are $1+a$ and $1-a$ for some value $a$ . Therefore, the sum is $\boxed{2}$

x+2y+z=x+y+1; if x0,y0 gives MAX then evidently -x0,-y0 gives MIN, so MIN+MAX=x0+y0+1-x0-y0+1=2