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Suppose we are studying the function $$ f(x,y) = xy + ax^2 + bx^2y^2, $$ We want to find the maximum $x$ satisfying the equation $$ f(x,y) = c, $$ where $a, b, c$ are constants. Somebody suggested to make use of the following auxiliary function $$ g(x,y) = xy, $$ so that $$ g + ax^2 + bg^2 = c. $$ Isolating $x$, $$ x^2 = \frac{c-g-bg^2}{a}. $$ Now he says that the same condition as $ \frac{dx}{dy} = 0 $ is

$$ \frac{dx^2}{d g(x,y)} = 0$$

Why?

Edit: I saw this trick here.

Daniele Tampieri
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tryst with freedom
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  • What's wrong with not using the substitution and rather let $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$ and then solving the system of equations simultaneously? – JayP Jul 17 '20 at 12:07
  • I saw this trick in another stack and I'm trying to understand it – tryst with freedom Jul 17 '20 at 12:07
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    I think its because that by applying the substitution $g(x,y)$, you turn the multivariable function into a function of a single variable. – JayP Jul 17 '20 at 12:13
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    Afaik the two conditions are only equivalent for $x \neq 0$. – Its_me Jul 24 '20 at 17:35
  • why is it equivalent? – tryst with freedom Jul 25 '20 at 07:34
  • If the curve can be split into pieces, where each of the variables $x$, $y$ and $g$ is a (bijective) function of each of the other (single) variables, then for $h=x^2$ we obtain $\frac{dh}{dy}=2x\frac{dx}{dy}=\frac{dh}{dg}\frac{dg}{dy}$ and for $x\neq 0$ and $\frac{dg}{dy}\neq 0$, thus, $\frac{dx}{dy} = 0 \Leftrightarrow \frac{dh}{dg} = 0$. You still have to check if it makes sense to exclude points that don't fulfill the two conditions. – Its_me Jul 25 '20 at 14:35
  • ... and for completeness if only one direction is required as in your example above, $\frac{dh}{dg} =0\Rightarrow \frac{dx}{dy}=0$ for all $x \neq 0$. – Its_me Jul 26 '20 at 11:17

1 Answers1

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I don't want to go into details of different cases, so I will assume that $x$ and $a,b,c$ are all positive, while $y$ is negative.

The canonical procedure would be to solve $f(x,y)=c$ with respect to $x$, to obtain $x=g(y)$, then solve $g'(y)=0$ to obtain $y_0$ (suppose it is unique), finally the solution would be $x_0 = g(y_0)$.

If we make a change of variables, as \begin{align} X &= x^2, \\ Y &= xy, \end{align} the equation $f(x,y)=c$ becomes $F(X,Y)=c$, i.e. $$ Y+aX+bY^2=c. $$ Taking into account that when $x$ is positive and maximum, also $X$ will be maximum, thanks the relation $X=x^2$ between them, then we can solve for $X$ and obtain $X=G(Y)$, i.e. $$ G(Y)=\frac{c-Y-bY^2}{a} $$ and solve $G'(Y)=0$ obtaining $$ Y_0 = -\frac{1}{2b} $$ and of consequence $$ X_0 = G(Y_0)=\frac{1}{a}\left(\frac{1}{4b}+c\right) $$ and finally $$ x_0 = \sqrt{\frac{1}{a}\left(\frac{1}{4b}+c\right)} $$

As a final remark, my $X=G(Y)$ corresponds to your $$ x^2 = \frac{c-g-bg^2}{a}. $$ while my $G'(Y)=0$ corresponds to your $$ \frac{dx^2}{d g(x,y)} = 0. $$

Vincenzo Tibullo
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