This question has been asked before but I want to lay out my method and get feedback on reasoning and process this took me a long to put together as I am new to the formatting:

Let the function $f$ be defined as $f$($x$,$y$,$z$) $=$ $x$$y$$z$

find the maximum and minimum values subject to the constraint: $g$($x$,$y$,$z$) $=$ $x^2$+2$y^2$+3$z^2$

Equation 1 $=$ $\nabla$$f_x$ $=$ $\lambda$ $\nabla$$g_x$ $=$ [ $yz$ $=$ $\lambda$2$x$]

Equation 2 $=$ $\nabla$$f_y$ $=$ $\lambda$ $\nabla$$g_y$ $=$ [ $xz$ $=$ $\lambda$4$y$]

Equation 3 $=$ $\nabla$$f_z$ $=$ $\lambda$ $\nabla$$g_z$ $=$ [ $yx$ $=$ $\lambda$6$z$]

It is my understanding that to solve these equations really just amounts to solving the relationship between values $x,y,z$ (and then plug into constraint) or the values themselves if there is an easy way to do so.

My basic instinct was to solve Equation 1 for $x$, obtaining: $x$ $=$ $\frac{yz}{2\lambda}$

Since Equation 3 consists of $x,y,z$ and I have x in terms of $y$ and $z$, then plugging $x$ into Equation 3 would get me the relationship of $y$ to $z$, giving me:

$\frac{yz}{2\lambda}$$\frac{y}{1}$ $=$ $\frac{\lambda6z}{1}$

Simplifying I then get: 12$z$$\lambda^2$ $=$ $y^2$$z$ Then Subtracting the right side over to the left and factoring out $z$ gives: $z$(12$\lambda^2$-$y^2$) $=$ 0

Giving me: $z$ $=$ 0 and 12$\lambda^2$ $-$ $y^2$ $=$ 0

$\lambda$ $\neq$ 0 because that would imply that $x=y=z=0$ which does not hold true in the constraint.

Now following from the above observation $z$ $\neq$ 0 because that would imply that $x=y=z=0$ which does not hold true in the constraint. This leaves 12$\lambda^2$ $-$ $y^2$ $=$ 0 for me to obtain values from.

Solving for $y$ I get: $y$ $=$ $\pm$ 2$\lambda$$\sqrt{3}$

I then plugged $y$ into the $x$ value I solved for in my first step: $x$ $=$ $\frac{yz}{2\lambda}$ and obtained:

$x$ $=$ $\pm$ $\sqrt{3}z$

From here I am lost. I can't seem to do anything to get the rest of the values and get numerical values. I would like my process addressed first and then suggestions, and by that I mean answering the question is there a way to proceed down the same route that I am going or was my process flawed from the start? The book gives the answer exactly as follows: maximum $\frac{2}{\sqrt{3}}$, minimum $\frac{-2}{\sqrt{3}}$. What does that even mean when checking my answers I am used to $f$($x,y,z$) $=$ $c$. I'm guessing that in this instance they are just giving me c.

Thank you

  • $\begingroup$ "find the maximum and minimum values subject to the constraint: $g$($x$,$y$,$z$) $=$ $x^2$+2$y^2$+3$z^2$" This is not a constraint, rather a definition of $g$. You might actually mean to ask: for some given $a>0$, find the maximum and minimum values of $f(x,y,z)$ subject to the constraint $g$($x$,$y$,$z$) $=a$, with $g(x,y,z)=$ $x^2$+2$y^2$+3$z^2$. $\endgroup$
    – Did
    May 19, 2016 at 5:54
  • $\begingroup$ This is somewhat similar to finding maximal possible volume of a parallelopiped in an ellipsoid. So looking at this post and other post which are linked there might perhaps help you. $\endgroup$ May 19, 2016 at 7:43

1 Answer 1


I have not been able to find the original question and I give you here my approach hoping it will help you.

We want to maiximize $xyz$ subject to the constraint $x^2+2 y^2+3 z^2=a$ with ($x\geq 0$) , ($y\geq 0$) , ($z \geq 0$).

So, let us consider $$F=x y z +\lambda \left(x^2+2 y^2+3 z^2-a\right)$$ Computing derivatives $$F'_x=y z+2 \lambda x=0\tag 1$$ $$F'_y=x z+4 \lambda y=0\tag 2$$ $$F'_z=x y+6 \lambda z=0\tag 3$$ $$F'_\lambda=x^2+2 y^2+3 z^2-a=0\tag 4$$ Now, I should consider equations $(1,2,3)$ and solve them for $x,y,z$ in terms of $\lambda$.

Using $(1)$ gives $x=-\frac{y z}{2 \lambda }$; plugging in $(2)$ leads to $y \left(4 \lambda -\frac{z^2}{2 \lambda }\right)=0$; plugging in $(3)$ leads to $z\left(6 \lambda -\frac{y^2}{2 \lambda }\right)=0$.

Considering all possibilities, the solutions of equations $(1,2,3)$ are then (hoping no mistakes) $$x=0\qquad y= 0\qquad z= 0$$ $$x= -2 \sqrt{6} \lambda \qquad y= -2 \sqrt{3} \lambda \qquad z=-2 \sqrt{2} \lambda$$

$$x= -2 \sqrt{6} \lambda \qquad y= +2 \sqrt{3} \lambda \qquad z=+ 2 \sqrt{2} \lambda$$ $$x= +2 \sqrt{6} \lambda \qquad y= -2 \sqrt{3} \lambda \qquad z= +2 \sqrt{2} \lambda$$

$$x= +2 \sqrt{6} \lambda \qquad y=+2 \sqrt{3} \lambda \qquad z= -2 \sqrt{2} \lambda$$ Using $(4)$, we just get $72 \lambda^2=a$, then $\lambda$ and the remaining follows.

  • $\begingroup$ Hmm your answers don't match the book though as they don't give the answers I stated. Do you know a different approach Claude. Thank you for taking the time to answer $\endgroup$
    – K. Gibson
    May 13, 2016 at 0:26
  • 1
    $\begingroup$ @K.Gibson: There is an error in the solution. Multiply (1), (2) and (3) by $x$, $y$ and $z$ respectively to get: $2 x^2=4 y^2=6 z^2$. Use these to substitute for $x^2$ and $y^2$ with the appropriate multiple of $z^2$ in (4) and solve for $z=\pm\sqrt{2/3}$ ... continue from there. $\endgroup$ May 14, 2016 at 5:41
  • $\begingroup$ Okay thank you that makes sense $\endgroup$
    – K. Gibson
    May 14, 2016 at 5:43
  • $\begingroup$ Okay so I just now got around to trying to work the problem out from what you said. Here is what I did. You said replace $x^2 and y^2$ with the appropriate multiples of z in 4 and solve. I found that $x$ was a multiple of $z$ by 3 and $y$ was a multiple of $z$ by $\frac{3}{2}$. Plugging them in I got 3+$\frac{3}{2}$+$3z^2-6$=0. My algebra lead me to $z$ = $\pm\frac{3}{\sqrt{2}}$ $\endgroup$
    – K. Gibson
    May 18, 2016 at 19:07
  • $\begingroup$ I got this by adding 6 over to the right then subtracted 3 leaving me: $\endgroup$
    – K. Gibson
    May 18, 2016 at 19:11

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