Find the maximum possible value of the expression
$\left(\frac{x^2 + 3y^2 + 9z^2}{1}\right) $
subject to
$x+2y +3z = 12$,
where $x,y,z$ are real numbers.
Find the maximum possible value of the expression
$\left(\frac{x^2 + 3y^2 + 9z^2}{1}\right) $
subject to
$x+2y +3z = 12$,
where $x,y,z$ are real numbers.
First a matter of terminology: the expression $\left(\frac{x^2+3y^2+9z^2}1\right)$ is not an equation: it does not say that two things are equal. It’s an expression, and your task is to maximize it subject to the condition (or constraint) that $x+2y+3z=12$. In this case the condition is that the point $\langle x,y,z\rangle$ lie in a particular plane, namely, the one whose equation is $x+2y+3z=12$.
Since you tagged this (algebra-precalculus), I’ll limit myself to purely algebraic techniques.
For convenience, if $p=\langle x,y,z\rangle$, let $f(p)=x^2+3y^2+9z^2$.
Consider only those points for which $z=0$, so that $x+2y=12$, and $x=12-2y$. For each real number $y$ let $p(y)=\langle 12-2y,y,0\rangle$; then $f\big(p(y)\big)=(12-2y)^2+3y^2=144-48y+7y^2=7\left(y-\frac{24}7\right)^2+\frac{432}7\;,$ where the last step was completing the square. Clearly this can be made arbitrarily large by taking $y$ large enough, so there is no point $p$ in the plane $x+2y+3z=12$ at which $f$ attains a maximum.
On the other hand, $f$ does have a minimum on $\pi$, and you can find it without calculus. Consider the points that lie in the plane $z=a$, where $a$ is any real number; their $x$- and $y$-coordinates must satisfy the equation $x+2y=12-3a$, or $x=12-3a-2y$.
Let $p_a(y)=\langle 12-3a-2y,y,a\rangle$ for each $y\in\Bbb{R}$. (The points $p(y)$ of the first part are $p_0(y)$ in this more general setting.) Then
$\begin{align*}f\big(p_a(y)\big)&=(12-3a-2y)^2+3y^2+9a^2\\ &=(12-3a)^2-4(12-3a)y+4y^2+3y^2+9a^2\\ &=7y^2-12(4-a)y+9(4-a)^2+9a^2\\ &=7\left(y^2-\frac{12}7(4-a)y\right)+9\left((4-a)^2+a^2\right)\\ &=7\left(\left(y-\frac67(4-a)\right)^2-\frac{36}{49}(4-a)^2\right)+18\left(8-4a+a^2\right)\\ &=7\left(y-\frac67(4-a)\right)^2-\frac{36}7(16-8a+a^2)+18\left(8-4a+a^2\right)\\ &=7\left(y-\frac67(4-a)\right)^2+\frac{18}7\left(24-12a+5a^2\right)\;, \end{align*}$
which clearly reaches its minimum when the first term is $0$, i.e., when $y=\frac67(4-a)$ and $f_a(y)=\frac{18}7\left(24-12a+5a^2\right)\;.\tag{1}$
Now we need only find the value of $a$ that makes $(1)$ as small as possible. Completing the square one more time, we see that
$\begin{align*} \frac{18}7\left(24-12a+5a^2\right)&=\frac{90}7\left(a^2-\frac{12}5a+\frac{24}5\right)\\ &=\frac{90}7\left(\left(a-\frac65\right)^2-\frac{36}{25}+\frac{24}5\right)\\ &=\frac{90}7\left(\left(a-\frac65\right)^2+\frac{84}{25}\right)\\ &=\frac{90}7\left(a-\frac65\right)^2+\frac{216}5\;, \end{align*}$
which reaches a minimum of $\dfrac{216}5$ at $a=\dfrac65$.
In other words, $f(p)$ is minimized when
$\begin{cases} z=\frac65\\\\ y=\frac67\left(4-\frac65\right)=\frac{12}{5}\\\\ x=12-3\left(\frac65\right)-2\left(\frac{84}{35}\right)=\frac{18}5\;, \end{cases}$
and its minimum value is $\dfrac{216}5$.