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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.

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    It is clear that there is no maximum, take for example $z=4$, $x=2M$, $y=-M$ where $M$ is huge. If we specify $x$, $y$, $z$ are all $\ge 0$, problem not too hard. For minimum use the Cauchy-Schwarz Inequality. Better I think is to let $x=3u$, $y=\sqrt{3}v$, $z=w$. Then looking at $9(u^2+v^2+w^2)$ subject to $3u+2\sqrt{3}v +3w=12$. Want nearest point to the origin on a plane, easy geometry or vectors argument. For max subject to $x$, $y$, $z$ all $\ge 0$, again not hard geometry problem.2012-03-06

1 Answers 1

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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 $\dfrac{216}5$.

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    @vikiiii: I can’t at the moment think of one that doesn’t require calculus. With first-semester calculus I could slightly shorten the calculations above, but the basic idea would remain the same. The simplest method, once you have the tools, is the method of [Lagrange multipliers](http://en.wikipedia.org/wiki/Lagrange_multiplier), which is accessible at least in straightforward problems once you have a little knowledge of partial derivatives.2012-03-06