Cfrgtkky

Let $f(x,y)=x(yln y-y)-yln x.$ Find $max_{1/2le xle 2}(min_{1/3le yle1}f(x,y))$.

This problem is quite easy and it is from Spivak; it is the part $c)$ of the general exercise 2-41 page 43 Calculus on manifolds; here it is:

Let $f:mathbb{R}times mathbb{R}to mathbb{R}$ be twice continuously differentiable. For each $xin mathbb{R}$ define $g_x(y)=f(x,y)$. Suppose that for each $x$ there is a unique $y$ with $g'_x(y)=0$; let $c(x)$ be this $y$.

$a)$: If $D_{2,2}f(x,y)ne0$ for all $(x,y)$ show that $c$ is differentiable and $c'(x)=-frac{D_{2,1}f(x,c(x))}{D_{2,2}f(x,c(x))}$

$b)$: Show that if $c'(x)=0$, then for some $y$ we have $D_{2,1}f(x,y)=0$, $D_2f(x,y)=0$.

I cannot visualize how part c) relates to the previous ones. Can you give me a hint?

I have given the complete solution below. The part highlighted shows where we use the result of part (a). In short, we use part (a) to compute the derivative of the critical point of $g_x(y)$ for the given function $f(x,y)$. This is used in computing the critical points to maximise $f(x,y)$ w.r.t. $x$ after minimising it w.r.t. $y$.

Let $f : (0,infty) times (0,infty) to mathbb{R}$ be the function given by $$f(x,y) = x(y log y - y) - y log x.$$
Clearly $f$ is $C^infty$. For each $x in (0,infty)$ define $g_x : (0,infty) to mathbb{R}$ by
$$g_x(y) = f(x,y).$$
To find points where $g_x$ is minimised, we first find the critical points:
$$
{g_x}'(y) = xlog y - log x,\
therefore {g_x}'(y) = 0 iff y = x^{1/x}.
$$
So, for each $x in (0,infty)$ there is a unique $y in (0,infty)$ such that ${g_x}'(y)=0$. So, let $c(x) = x^{1/x}$ be this critical point. To check the nature of this critical point, we evaluate ${g_x}''(c(x))$ and check its sign.
$$
{g_x}''(y) = frac{x}{y} implies {g_x}''(c(x)) = frac{x}{x^{1/x}}.
$$
Hence, ${g_x}''(c(x)) > 0$ for all $x in (0,infty)$, so $g_x(y)$ has a global minimum at $y=c(x)$. Therefore, we would like to conclude that
$$
min_{1/3 leq y leq 1} { f(x,y) } = f(x,c(x)) = x cdot x^{1/x},
$$
but this would be a bit hasty, for it is not necessary that $c(x) in [1/3,1]$ for all $x in (0,infty)$. So, let $alpha in (0,infty)$ be the unique element such that $c(alpha) = 1/3$. Define $h : (0,infty) to mathbb{R}$ by
$$
h(x) =
begin{cases}
f(x,1/3), & x in (0,alpha);\
f(x,c(x)), & x in [alpha,1];\
f(x,1), & x in (1,infty)
end{cases}
=
begin{cases}
-(x(1+log 3)+log x)/3, & x in (0,alpha);\
-x cdot x^{1/x}, & x in [alpha,1];\
-x-log x, & x in (1,infty).
end{cases}
$$
Then,
$$
min_{ 1/3 leq y leq 1 } { f(x,y) } = h(x).
$$

Now, note that the hypothesis of part (a) of the problem is satisfied, because $D_{2,2} f(x,y) = {g_x}''(y) neq 0$ for all $x,y in (0,infty)$. Hence,
$$
D_{2,1} f(x,y) = log y - frac{1}{x},
$$
and
$$
begin{align}
c'(x) = -frac{D_{2,1}f(x,c(x))}{D_{2,2}f(x,c(x))}
= frac{x^{1/x}(1 - log x)}{x^2}.
end{align}
$$

To find points where $h$ is maximised, we find its critical points. In the interval $(0,alpha)$,
$$
h'(x) = -frac{left(1+log 3 + frac{1}{x}right)}{3}
$$
which is negative for all $x in (0,alpha)$. Hence, $h(x)$ is decreasing on this interval. In the interval $(1,infty)$,
$$
h'(x) = -1-frac{1}{x},
$$
which is negative for all $x in (1,infty)$. Hence, $h(x)$ is decreasing on this interval as well. In the interval $[alpha,1]$,
$$
h'(x) = -c(x) - x c'(x) = -c(x)left( 1 + frac{1-log x}{x} right),
$$
which is negative for all $x in [alpha,1]$. Hence, $h$ is decreasing on this interval as well. One can check that $h(x)$ is continuous at $x = alpha$ and $x = 1$, so $h$ is continuous everywhere, and thus it is decreasing on the entire domain.
Lastly, $c(1/2) = 1/4 < 1/3$, so $1/2 < alpha$. Therefore, we have
$$
begin{align}
max_{1/2 leq x leq 2} left{ min_{1/3 leq y leq 1} { f(x,y)} right} &= max_{1/2 leq x leq 2} { h(x) }\
&= h(1/2)\
&= f(1/2,1/3)\
&= -frac{left( 1-log frac{3}{4} right)}{6}.
end{align}
$$