Minimum Hui

We can rewrite the given function by:

$F=\frac{1}{3}(2x-3y)^2+\frac{1}{5}(5y-6z)^2+\frac{1}{15}(9z-5x)^2$

$+4(y-1)^2+\frac{2}{5}(6z-5)^2+86$ .

It is clear that $F \geq 0$ . $F=0$ when $2x-3y=5y-6z=9z-5x=y-1=6z-5=0$ .

These equations lead to $x=\frac{3}{2}$ , $y=1$ , $z=\frac{5}{6}$ .

Therefore, the minimum value of F is 86, where $x=\frac{3}{2}$ , $y=1$ , $z=\frac{5}{6}$ .

Let $f(x,y,z) = 3x^2+12y^2+27z^2-4xy-12yz-6xz-8y-24z+100$ We are given that all variables are real numbers and aren't restricted in any way. Hence consider the system $\begin{aligned} \frac{\partial f}{\partial x} = 6x - 4y - 6z \\ \frac{\partial f}{\partial y} = -4x + 24y - 12z - 8 \\ \frac{\partial f}{\partial z} = -6x - 12y + 54z - 24 \end{aligned}$ This system has solution $(x,y,z) = \left(\frac{3}{2},1,\frac{5}{6} \right)$ . Hence the minimum is $f\left(\frac{3}{2},1,\frac{5}{6}\right) = \boxed{86}$ The fact that this is not maximum is obvious, the function $f(x,y,z)$ approaches infinity for $x$ approaching infinity and $y,z$ zero. It could also be saddle point, but I won't discuss this here.

Train of thoughts: Basically I attempted to factorize , although the expression did not look symmetric, which came out as nothing nicer than: $x (3 x-4 y-6 z)+y (12 y-12 z-8)+z (27 z-24)+100$ . Maybe I missed something, but I gave up and went on to consider the polynomial discriminant , still in an attempt to learn more about the roots, which was obviously useless. Perhaps, I'm missing something trivial, but when one sees ugly powers and one wants to bring down the exponents , one way to do so is to consider calculus, i.e. derivatives . And clearly this expression when plotted is a concave up parabola, so calculus indeed works.

We first consider the derivative of the expression and set it to $0$ to consider the relationship between $x,y,z$ . Computing the derivative, we have:

$\frac{d}{dx} 3x^2+12y^2+27z^2-4xy-12yz-6xz-8y-24z+100$

$= 3 \frac{d}{dx}(x^2) - 4y\frac{d}{dx}(x) - 6z\frac{d}{dx}(x) + \frac{d}{dx}(12y^2) +\frac{d}{dx}(27z^2) + \frac{d}{dt}(-12yz) + \frac{d}{dt}(-8y) +\frac{d}{dt}(-24z) + \frac{d}{dx}(100)$ (where we've differentiated each term and pulled out the constants)

$= 6x - 4y -6z$ (recall that $\frac{d}{dt}(y) = 0$ and $\frac{d}{dt}(z)$ = 0, etc. and the power rule)

Which we set to $0$ , subbing $6x = 4y - 6z$ back into the big expression, we get:

$4 (6 x^2-6 x (2 y+1)+8 y^2+2 y+25) (*)$

That's the beauty of taking derivatives, we eliminate the variables one by one . Once again, and this time for simplicity sake, we take the derivative of $6 x^2-6 x (2 y+1)+8 y^2+2 y+25 (**)$ and set it to $0$ :

$\frac{d}{dt} (6 x^2-6 x (2 y+1)+8 y^2+2 y+25) = 6\frac{d}{dt}(x^2) - 6(2y+1)\frac{d}{dt}(x) + \frac{d}{dt}(8y^2) + \frac{d}{dt}(2y) + \frac{d}{dt}(25)$

$= 12x - 6(1+2y) = 6(2x - 2y - 1)$

Now, $6(2x - 2y - 1) = 0 ⇔ 2x = 2y + 1$ , subbing this into $(**)$ , we get:

$2y^2 -4y + \frac{47}{2}$ .

With one variable, it is simple to complete the square , which then becomes: $2 (y-1)^2+\frac{43}{2}$ . Clearly the minimum is $\frac{43}{2}$ . Recall that in $(*)$ , we have to times $4$ , so the final answer is $4 \times \frac{43}{2}$ , which evaluates as 86 .

We will prove that the minimum value is 86. Applying AM-GM inequality yields: $\frac{4x^2}{3}+3y^2 \ge 4xy$ And $5y^2+\frac{36z^2}{5} \ge 12yz$ And $\frac{27z^2}{5}+\frac{5x^2}{3} \ge 6zx$ We also have: $4(y-1)^2 \ge 0$ And $\frac{2}{5}(6z-5)^2 \ge 0$ In conclusion, we have the minimum of the function is 86.

Arrange the given expression, we get

$2(x-y)^2+(x-3z)^2+6(y-z)^2+4(y-1)^2+12(z-1)^2+84$

Take $(y,z)=(1,1)$ , because we will get three terms of $0$ .

The two terms left, let it be $A$ , are

$2(x-y)^2+(x-3z)^2$

$=3x^2+(4y+6z)x+(2y^2+9z^2)$

$\displaystyle =3\bigg (x-\frac{2y+3z}{3}\bigg )^2+\bigg (\frac{2}{3}y^2-4yz+6z^2\bigg )$

The minimum value of $A$ is

$\displaystyle \frac{2}{3}y^2-4yz+6z^2$ .

When $(y,z)=(1,1)$ , the value of $A$ is $2\frac{2}{3}$ , the minimum value of the expression is $86\frac{2}{3}$ .

If we want to reduce the minimum value of $A$ , we can either reduce $y$ or $z$

By analysing, reduce the value of $z$ is a better approach, so let $y=1$ , $z$ is reduced,

The value of $A$ will be reduced by $2-(6z^2-4z)$ ,

The increase of the other terms will be $18(z-1)^2$ ,

The overall reduced value will be $2-(6z^2-4z)-18(z-1)^2$ .

Now, the maximum value that can be reduced,

$\max\bigg [2-(6z^2-4z)-18zy-1)^2\bigg ]=\frac{2}{3}$

Hence, the minimum value of the expression is $86\frac{2}{3}-\frac{2}{3}=\boxed{86}$

To get the minimum value of a function in x, we find the derivative of the function, equate it to 0 and solve for x, then substitute the answer to the function. The extension of this concept to a function of three variables (in x, y and z) entails us to find the partial derivative of the expression with respect to x, y and z, equate all of them to 0, then solve the resulting system of equations. The answers are then substituted to the original expression.

Let $f(x,y,z) = 3x^2 + 12y^2 + 27z^2 - 4xy - 12yz - 6xz - 8y - 24z + 100$ .

Then $f_{x}(x,y,z) = 6x - 4y - 6z$ .

We also have $f_{y}(x,y,z) = 24y - 4x -12z - 8$ .

Finally $f_{z}(x,y,z) = 54z - 12y - 6x - 24$ .

Now we equate each of them to 0. We have the ff. system of equations:

$6x - 4y - 6z = 0$ ------------ (1)

$-4x + 24y - 12z = 8$ ------------ (2)

$-6x - 12y + 54z = 24$ ------------ (3)

We add (1) to (3) to get

$-16y + 48z = 24$ ------------ (4)

We multiply both sides of (2) by $1.5$ to get

$-6x + 36y - 18z = 12$ ------------ (5)

Now we add (1) to (5) to get

$32y - 24z = 12$ ------------ (6)

Now we divide both sides of (6) by 2 to get

$16y - 12z = 6$ ------------ (7)

We add (4) to (7) to get 36z = 30). Then \(z = \frac {5}{6} .

Now we substitute the value that we got for z to equation (7) to get

$16y - 12(\frac {5}{6}) = 6$ .

Solving for y, we have $16y - 10 = 6$ , $16y = 16$ and $y = 1$ . Finally we substitute the known values of y and z to equation (1) to get

6x - 4(1) - 6(\frac {5}(6}) = 0 .

Solving for x, we get $6x - 4 - 5 = 0$ , $6x - 9 = 0$ and $x = \frac {3}{2}$ .

Now we have the values of x, y and z, we substitute each to the original expression. We have:

$3x^2 + 12y^2 + 27z^2 - 4xy - 12yz - 6xz - 8y - 24z + 100$

$= 3(\frac {3}{2})^2 + 12(1)^2 + 27(\frac {5}{6})^2 - 4(\frac {3}{2})(1) - 12(1)(\frac {5}{6}) - 6(\frac {3}{2})(\frac {5}{6}) - 8(1) - 24(\frac {5}{6}) + 100$

$= 3(\frac {9}{4}) + 12 + 27(\frac {25}{36}) - 6 - 10 - \frac {15}{2} - 8 - 20 + 100$

$= \frac {27}{4} + 68 + \frac {75}{4} - \frac {30}{4}$

$= \frac {72}{4} + 68$

$= 18 + 68$

$= 86$

let f(x,y,z) equal the polynomial above, take partial derivatives by x, y and z to obtain the following equations

6x - 4y - 6z = 0 24y - 4x - 12z = 8 54z - 12y - 6x = 24

solve for x,y,z and check that it is a minimum

For single variable functions such as $f(x)$ , we get the maximum/minimum by getting the first order derivative and equating it to zero. In cases such as this, we have multiple variables involved. $f(x,y,z)$ can be partially differentiated by considering $y$ and $z$ as constants.

$\frac{\partial f}{\partial x} (x,y,z) = \frac{\partial}{\partial x}(3x^2-4xy-6xz)$

To shorten the process, I have removed all terms independent of $x$ { $12y^2, 27z^2, -12yz, -8y, -24z, 100$ }. We find the first partial derivative to be $6x - 4y - 6z$ . Notice that this is incomplete for we can only get the minimum of $x$ if we have values for $y$ and $z$ , also at their minimum. Thus, we must get the partial derivatives in terms of $y$ and $z$ as well.

$\frac{\partial f}{\partial y} (x,y,z) = \frac{\partial}{\partial y}(12y^2-4xy-12yz-8y)$ $\frac{\partial f}{\partial y} (x,y,z) = 24y - 4x -12z - 8$

Again, we disregard all terms invariable with $y$ , this time holding $x$ and $z$ constant. Repeating this process, we get the derivative of $f(x,y,z)$ in terms of $z$ . We get $54z - 12y - 6x - 24$ .

Equating everything to $0$ , we get the minimum functions for $x$ , $y$ and $z$ , respectively.

1. 6x - 4y - 6z = 0
2. 24y - 4x -12z - 8 = 0
3. 54z - 12y - 6x - 24 = 0

We can then proceed to solve for each. The solution is $(\frac{3}{2},1,\frac{5}{6})$ - the minimum values for x, y and z. We can then use these to find the minimum value for the original function to get $86$ .

The expression in the question can be rewritten as (2x-3y)^2/3+(5x-9z)^2/15+(5y-6z)^2/15+4(y-1)^2+2(6z-5)^2/5+86. So the minimum must be 86 achieved at (3/2, 1, 5/6).

define f as a function in x,y,z. solve the linear system: df/dx=0 df/dy=0 df/dz=0

f(x,y,z)=86

We differentiate the given expression with respect to $x,y,z$ and get $6x-4y-6z, -4x+24y-12z-8, -6x-12y+54z-24$ respectively. It is easy to see all three derivatives have their second-order derivative positive, which means that there is a global minimum. So we let $6x-4y-6z=0, -4x+24y-12z-8=0, -6x-12y+54z-24=0$ , to find where the minimum is. Adding the first and third equalities gives $6z-2y=3$ . Multiply the first by 2 and the second by 3 then add them up gives us $8y-6z=3$ . Adding up these 2 equations give $y=1$ . Substituting this into other equation gives us the value of $x,z=\frac{3}{2}, \frac{5}{6}$ and locate the global minimum. Substituting these to the original expression gives 86.

Do partial differentiation of the equation to obtain 3 equations as follow and equate them them with 0,

$3x-2y-2z=0$ (simplified) $-x+6y -3z-2=0$ $-x-2y+9z-4=0$

Solving this we get the equation to be minima $(x,y,z)=(\frac{3}{2}, 1, \frac{5}{6})$

Substituting this into original equation we get minima to be 86.

The analytic solution:

Define the following function: f = 3 x^2 + 12 y^2 + 27 z^2 - 4 x y - 12 y z - 6 x z - 8 y - 24*z + 100 Then the minimum occurs for solving $\nabla f=0$ : Solve[{D[f, x] == 0, D[f, y] == 0, D[f, z] == 0}]

Of course one could derive and solve the system by hand, but since this is trivial, it its left out.

The numerical solution: FindMinimum[{3 x^2 + 12 y^2 + 27 z^2 - 4 x y - 12 y z - 6 x z - 8 y - 24*z + 100}, {{x, 1}, {y, 1}, {z, 1}}]