• Method 1:

Note that $x^{2} + y^{2} + z^{2} = (x + y + z)^{2} - 2(xy + yz + xz) = 15^{2} - 2 \times 72 = 81$ , which describes a sphere of radius $9$ centered at the origin. So the solution set to the given set of equations will be the intersection of this sphere with the plane $x + y + z = 15$ , which describes a circle. This circle is symmetric about the plane $y = z$ , (as $y$ and $z$ are interchangeable in the original system of equations), so both the maximum and minimum of $x$ will occur when $y = z$ . This transforms the given set of equations to

$x + 2y = 15 \Longrightarrow 2y = 15 - x$ and

$2xy + y^{2} = y(2x + y) = 72 \Longrightarrow 2y(4x + 2y) = 288 \Longrightarrow (15 - x)(4x + (15 - x)) = 288 \Longrightarrow$

$(15 - x)(3x + 15) = 288 \Longrightarrow (15 - x)(x + 5) = 96 \Longrightarrow x^{2} - 10x + 21 = (x - 3)(x - 7) = 0$ .

So $S = 3$ and $M = 7$ , and thus $S + M = \boxed{10}$ .

The minimum for $x$ is achieved when $(x,y,z) = (3,6,6)$ and then maximum when $(x,y,z) = (7,4,4)$ .

• Method 2:

Note that $y + z = 15 - x$ and $yz = 72 - x(y + z) = 72 - x(15 - x) = 72 - 15x + x^{2}$ . Then

$0 \le (y - z)^{2} = (y + z)^{2} - 4yz = (15 - x)^{2} - 4(72 - 15x + x^{2}) \Longrightarrow$

$0 \le 225 - 30x + x^{2} - 288 + 60x - 4x^{2} = -3x^{2} + 30x - 63 = -3(x^{2} - 10x + 21) \Longrightarrow$

$0 \le -(x - 3)(x - 7) \Longrightarrow 3 \le x \le 7$ ,

which is in agreement with the previous result.

To find the domain, x, we will have to substitute a variable to make a 2-variable equation. We take the first equation and put it in terms of z: $x+y+z = 15 \implies z=15-x-y$ Plugging this into the second equation, we get $xy + y(15-x-y) + x(15-x-y) = 72 \\ xy+15y-xy-y^2+15x-x^2-xy=72 \\ -y^2+(15-x)y+15x-x^2-72=0 \\ y^2+(x-15)y+x^2-15x+72=0$ Using the quadratic formula, we get $y = \frac{-(x-15) \pm \sqrt{x^2-30x+225-4(x^2-15x+72)}}{2} \\ y= \frac{15-x \pm \sqrt{-3x^2+30x-63}}{2}$ Knowing that y is a real number and not an imaginary number, we know $-3x^2+30x-63$ must be positive. Therefore, we can now set boundaries and find the maximum and minimum values. $-3x^2+30x-63>0 \\ 3x^2-30x+63<0 \\ 3x^2-30x+63=0 \\ x = \frac{30 \pm \sqrt{900-4(3)(36)}}{2 \cdot 3} \\ S = \frac{30-\sqrt{900-432}}{6}, M = \frac{30+\sqrt{900-432}}{6} \\ S+M=5+5+\frac{\sqrt{468}}{6}-\frac{\sqrt{468}}{6} = \boxed{10}$

My method involves the use of Cauchy-Schwarz Inequality.

We manipulate the equations such that we can do the inequality in terms of $x$ .

$x+y+z=15 \\ y+z=15-x$

$xy+yz+zx=72 \\xy+xz=72-yz \\ x(y+z)=72-yz \\ x(15-x)=72-yz$

$\Rightarrow yz=x^2-15x+72$

$y^2+z^2=(15-x)^2-2(x^2-15x+72)=81-x^2$

$(y^2+z^2)(1+1)\geq (y+z)^2$ (Cauchy Schwarz Inequality)

$2(81-x^2)\geq (15-x)^2$

$3x^2-30x+63 \leq 0$

$3(x-7)(x-3) \leq 0$

So $3\leq x\leq 7$

$\Rightarrow S=3, M=7, S+M=10$

Let $x, y$ and $z$ be the solutions of a cubic equation. Then the equation $p^3 - 15p^2 + 72p - c$ would have it's roots as $x, y$ and $z$ . Then $p^3 - 15p^2 + 72p = c$ would have three solutions if c $\in$ (108,112) as c would be a horizontal line parallel to the x-axis. All the three solutions are symmetrical so the smallest valued solution would be when c = 108, giving us two equal roots 6 and the other root 3 (Sum of the roots is 15). The largest valued solution would occur when c = 112, giving us two equal roots 4 and the other root 7.

So, $S = 3, M = 7$ giving us $\boxed{10}\$

First I transform to a new set of variables: $a = x - 5$ , $b = y - 5$ , $c = z - 5$ . Then $\begin{cases} a + b + c = 0 \\ ab + bc + ac = -3\end{cases}$ Suppose we have a solution $(a,b,c)$ for this system. Then $(-a,-b,-c)$ is obviously also a solution. In particular, if $(a^\star,b^\star,c^\star)$ is a solution with minimum $a$ , then $(-a^\star,-b^\star,-c^\star)$ is a solution with maximum $a$ and vice versa.

Therefore the answer to the problem is $x_{\text{min}} + x_{\text{max}} = (5 + a_{\text{min}}) + (5 + a_{\text{max}}) = (5 + a^\star) + (5 - a^\star) = 5 + 5 = 10.$

In general, if $\begin{cases} x + y + z = 3p \\ xy + yz + zx = 3q \end{cases}$ we have $x_{\text{min}} + x_{\text{max}} = 2p.$

With our transformed variables $(a,b,c) = (x-5,y-5,z-5)$ it is also easier to find the minimum and maximum solutions.

Substitute $c = -a-b$ : $ab + bc + ac = -3\ \ \ \ \therefore \ \ \ \ a^2 + b^2 + ab = 3.$ To find extreme values of $a$ , differentiate and set $da = 0$ : $\begin{cases} (2a + b)da + (2b + a)db = 0 \\ da = 0 \end{cases} \ \ \ \ \therefore\ \ \ \ 2b + a = 0 \ \ \ \ \therefore\ \ \ \ a = -2b.$ $\therefore\ \ \ \ (-2b)^2 + b^2 + (-2b)b = 3 \ \ \ \ \therefore\ \ \ \ 3b^2 = 3 \ \ \ \ \therefore\ \ \ \ b = \pm 1$ $\therefore\ \ \ \ \begin{cases} (a,b,c) = (-2,1,1) \\ \text{or} \\ (a,b,c) = (2,-1,-1) \end{cases} \ \ \ \ \therefore\ \ \ \ \begin{cases} (x,y,z) = (3,6,6) \\ \text{or} \\ (x,y,z) = (7,4,4) \end{cases}.$ Thus we submit the solution $x_{\text{min}} + x_{\text{max}} = 3 + 7 = \boxed{10}$ .

The general minimum and maximum solutions for $\begin{cases} x + y + z = 3p \\ xy + yz + zx = 3q \end{cases}$ are $x_{\text{min, max}} = p \mp 2\sqrt{p^2 - q}.$

I used the method of Lagrangian multipliers.

We have $x+y+z=15,$ and $xy+yz+zx=72.$ The 2nd eq $\implies x= \frac{72-yz}{y+z}.$

So maximizing/minimizing $x$ means maximizing/minimizing $x= \frac{72-yz}{y+z}.$ The constraint being $x+y+z=15.$

$\therefore$ using Lagrangian multiplier, $\nabla{(x+y+z-15)} = \lambda \nabla{( \frac{72-yz}{y+z})} , \text{where} \nabla = ( \frac{\partial}{\partial{x}} , \frac{\partial}{\partial{y}} , \frac{\partial}{\partial{z}} ) .$

Differentiating and comparing both sides, we get $\lambda =0,$ and so $y=z.$ Substituting this into the given equations and solving for x, we get a quadratic. Turns out that $x= 3$ or $x=7$ .

Considering P(q) = q³ + bq² +cq +d with roots x, y and z.

x+y+z = 15 implies that b = -15 and xy +yz +xz = 72 implies that c= 72.

So P(q) = q³ -15q² +72q +d.

Deriving this polynomial we will find the maximum and minimum values.

P'(q) = 3q² -30q +72 =0
Solving this equation, we can find a minimum value of 4 and a maximum value of 6

if these values are x values the sum is 10, if they're not, we have

z= 4 => x+y = 11 and xy = 28 => x = 7 and y =4 => (x,y,z) = (7,4,4)

z= 6 => x+y = 9 and xy = 18 => x=3 and y =6 => (x,y,z) = (3,6,6)

Code

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for x in range(1,100):
    for y in range(0,100):
        for z in range(0,100):
            if (x+y+z==15):
                if (x*y+y*z+x*z==72):
                    print(x)

Output

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Answer

$\text{Min+Max}=3+7=\boxed{10}$

The way i thought of is as:

since x, y ,z are real numbers. once of them can be zero, say y = 0, then i will be left with x + z = 15 and xz = 72

possible values for xz=72 can be 1 x 72 2 x 36 3 x 24 4 x 18 6 x 12 8 x 9 and for x+z = 15, the closest possible values can be: 8+9 = 17 or 6 + 12 = 18 which are still greater then 15, so, 5 <= x <= 8. so with kind of a wild guess, since i had 3 attempts, i started with 10 as an option and it was correct,

this is was just a thought process not a solution ;)