For real numbers $x, y$ in the interval $[0 ; 12]$:

$$
x y=(12-x)^{2}(12-y)^{2} \text {. }
$$

What is the largest value of the product $x y$?
I. solution. Non-negative numbers stand on both sides of the equation, so we can take the square root. Since $x$ and $y$ are in the interval $[0 ; 12]$, the square root of the right side is $(12-x)(12-y)=12^{2}-12(x+y)+x y$. Thus, we get

$$
\sqrt{x y}=12^{2}-12(x+y)+x y=12^{2}-24 \frac{x+y}{2}+x y
$$

The right side does not decrease if we replace the arithmetic mean of $x$ and $y$ with their geometric mean, which is not greater:

$$
\sqrt{x y} \leq 12^{2}-24 \sqrt{x y}+x y
$$

Substituting the variable $t=\sqrt{x y}$, we get the quadratic inequality

$$
0 \leq 12^{2}-25 t+t^{2}=(t-9)(t-16)
$$

Since

$$
0 \leq t=\sqrt{x y} \leq \frac{x+y}{2} \leq 12
$$

the variable $t$ also falls into the interval $[0 ; 12]$. On this set, the solution to the inequality $0 \leq (t-9)(t-16)$ is the interval $[0 ; 9]$. This implies that for the variables of the problem, $\sqrt{x y} \leq 9$, hence $x y \leq 81$.

Equality is possible, indeed, when $\sqrt{x y}=9$, the conditions of the problem are satisfied if and only if (1) holds, that is,

$$
9=12^{2}-24 \frac{x+y}{2}+81, \quad \text { so } \quad \frac{x+y}{2}=9
$$

The value of the product being examined can thus be 81, if the arithmetic and geometric means of the variables are both 9, that is, $x=y=9$.

Therefore, under the given conditions, 81 is the maximum value of $x y$.

II. solution. The equation is the equation of a set of points $H$ in a rectangular coordinate system. The set $H$ is contained in the 12 unit side square whose one vertex is the origin, and whose sides lie on the positive parts of the coordinate axes. We need to find the maximum value of the product $x y$ on this set, which is a so-called conditional extremum problem.

We prove that for any point $P(x ; y)$ in the set $H$,

$$
x+y \leq 18
$$

From this, the answer to the problem's question immediately follows: by the inequality of arithmetic and geometric means,

$$
x y \leq\left(\frac{x+y}{2}\right)^{2} \leq 9^{2}=81
$$

and if $x=y=9$, then equality holds in both inequalities; the maximum value of the product $x y$ is 81.

It is clearly sufficient to prove (2) in the case when $x \geq 6$, since $y \leq 12$. Our statement can be rearranged in two ways: $y \leq 18-x$, or $x-6 \leq 12-y$, and indeed, according to our remark, the latter can also be squared. This means that either $y \leq 18-x$, that is,

$$
x y \leq x(18-x)
$$

or $(x-6)^{2} \leq(12-y)^{2}$, that is,

$$
(12-x)^{2}(x-6)^{2} \leq(12-x)^{2}(12-y)^{2}
$$

holds, from which the statement to be proved follows. On the set $H$, the left side of (3) is equal to the right side of (4), so it is enough to show that if $6 \leq x \leq 12$, then

$$
(12-x)^{2}(x-6)^{2} \leq x(18-x)
$$

Let $s=x-9$. Then $s \in[-3 ; 3]$ and we need to prove that

$$
(s-3)^{2}(s+3)^{2} \leq(s+9)(9-s)
$$

Performing the operations and rearranging:

$$
s^{4}-17 s^{2}=s^{2}\left(s^{2}-17\right) \leq 0
$$

The second factor is negative on the interval $[-3 ; 3]$, and the first is not. We have proved statement (5), thus completing the solution.

Remark. The set $H$ is shown in the figure. The curve is clearly symmetric about the line $y=x$. The statements of the above proof can be read from the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_40423e7ce0efbf9654c6g-2.jpg?height=575&width=585&top_left_y=130&top_left_x=759)

III. solution. We prove that if the numbers $x, y$ in the interval $[0 ; 12]$ satisfy the equality of the problem, then $x y \leq 81$. This implies that the maximum value of the product is 81, since with the choice $x=y=9$, equality holds everywhere.

The proof is indirect. Suppose there is a pair of numbers $x, y$ such that $0 \leq x, y \leq 12$ and

$$
(12-x)^{2}(12-y)^{2}=x y>81
$$

Note that in this case $0<12-x \leq 12$ and $0<12-y \leq 12$, so

$$
12-\frac{9}{12-x}>0 \quad \text{and} \quad 12-\frac{9}{12-y}>0
$$

Both factors are positive, so it follows that

$$
12-\frac{9}{12-x}>y
$$

Multiplying by the positive $x$ and using the indirect assumption again,

$$
x\left(12-\frac{9}{12-x}\right)>x y>81
$$

Multiplying the last inequality by the positive $(12-x)$ and rearranging, we get

$$
-4 x^{2}+72 x-18^{2}=-(2 x-18)^{2}>0
$$

which is clearly impossible. We have completed the proof, and under the given conditions, the product $x y$ indeed cannot take a value greater than 81.