Multivariable calculus max/min

Question

Find and classify the critical points of this function: $f(x,y)= (x^y)-(xy)$ in the domain $x>0, y>0$.

I am having trouble treating x and y as constants when taking partial derivatives.

Answer 1

Hint

If the names make trouble to you because they are names of variables, consider first $f_1(x)=x^k-kx$ and compute its derivative with respect to $x$ as usual; this will give you $$\frac {df_1(x)}{dx}=k x^{k-1}-k$$ so, replacing $k$ by $y$, $$f'_x(x,y)=y x^{y-1}-y$$ Do a similar thing using $f_2(y)=k^y-ky$; this will give you $$\frac {df_2(y)}{dy}=k^y \log(k)-k$$ so, replacing now $k$ by $x$ $$f'_y(x,y)=x^y \log (x)-x$$

Is this making things clearer ? When you take the partial derivative with respect to one variable the other variables are considered as constants.

Answer 2

Claude Leibovici approach is correct and elegant. However in the general case, you can avoid treating variables as constants, by using derivatives rather than differentials.

A derivative is an operator d such that $d(f+g) = d(f) + d(g) = df + dg$, and $d(fg) = f d(g) + g d(f) = f dg+g df$. For the derivative, you further have $df(u) = f'(u) du$.

For example: $de^u = e^u du$, $d\log(u) = \frac{du}{u}$, and $$\begin{align*} d(x^y) &= d(e^{y \log x}) = e^{y \log x} d(y \log x) = x^y [\log x dy+y d(\log x)] = x^y [\log x dy+y \frac {dx} x]\\ &= (y x^{y-1}) dx + (x^y \log x)dy. \end{align*}$$

This is coherent with $\frac {\partial x^y} {\partial x} = y x^{y-1}$ and $\frac {\partial x^y} {\partial y} = x^y \log x$.

In your case (and slowly by slowly): $$\begin{align*} df &= d(x^y-xy) = d(x^y) - d(xy) \\ &= (y x^{y-1}) dx + (x^y \log x)dy - xdy - ydx \\ & = (y x^{y-1} -y)dx + (x^y\log x -x)dy. \end{align*}$$

which is coherent with Claude's results.

On the critical points, the derivative vanishes: $$\begin{cases} y \cdot x^{y-1} - y &= 0\\ x^y\log x -x &= 0 \end{cases}$$

The first line becomes $y=0$ or $x^{y-1}=1$, that is $y=0$, $x=0$ or $y-1=0$. Only $y=1$ is in the domain $x>0, y>0$.

Putting $y=1$ in the second line, gives $x\log x = x$, which is true for $x=0$ or $x=e$.

Hence, there is one critical points $(x,y)=(e,1)$. Compute $f(e,1)$ to see if it is a maximum or a minimum.