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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.

share|cite|improve this question
where is the denominator? – Riccardo.Alestra Mar 6 '12 at 7:44
Wrote the denominator. – vikiiii Mar 6 '12 at 8:10
As @Riccardo.Alestra writes, the denominator is missing. If it is 1, the question doesn't make sense, since then there is no maximal value, as the values on $(-t,-t,4+2t)$ tend to infinity as $t \to \infty$. – martini Mar 6 '12 at 8:11
@martini your answer is right.But how did you get (−t,−t,4+2t) – vikiiii Mar 6 '12 at 8:14
I solved $x+2y+3z = 12$ and did wrong, I meant $(-t, -t, 4+t)$ ... sorry. – martini Mar 6 '12 at 8:16
up vote 1 down vote accepted

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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Thanks Brian. But have a question.If i have not tagged this as (algebra-precalculus) , is there is any other easy way to answer the question. – vikiiii Mar 6 '12 at 11:33
@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, which is accessible at least in straightforward problems once you have a little knowledge of partial derivatives. – Brian M. Scott Mar 6 '12 at 11:44

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