# Find the maximum possible value of an expression, subject to a linear constraint

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.

• where is the denominator? Commented Mar 6, 2012 at 7:44
• Wrote the denominator. Commented Mar 6, 2012 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 $t \to \infty$. Commented Mar 6, 2012 at 8:11
• @martini your answer is right.But how did you get (−t,−t,4+2t) Commented Mar 6, 2012 at 8:14
• I solved $x+2y+3z = 12$ and did wrong, I meant $(-t, -t, 4+t)$ ... sorry. Commented Mar 6, 2012 at 8:16

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 $\dfrac{216}5$.

• 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. Commented Mar 6, 2012 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. Commented Mar 6, 2012 at 11:44