Question

Let $(x, y, z)$ be an ordered triplet of real numbers, such that $x < 1, y < 2, z < 3$ and $x + \frac{y}{2} + \frac{z}{3} > 0$. Let $(x_0, y_0, z_0)$ be the triplet $(x, y, z)$ for which expression $(1 - x)(2 - y)(3 - z)(x + \frac{y}{2} + \frac{z}{3})$ is maximum, then the value of $\frac{z_0}{x_0y_0}$ is

• A. 0

Answer 1

6

Answer 2

To maximize the expression $P = (1 - x)(2 - y)(3 - z)(x + \frac{y}{2} + \frac{z}{3})$ under the given constraints, we use the AM-GM inequality.

1. Introduce new variables: Let $a = 1 - x$, $b = 2 - y$, $c = 3 - z$. From the constraints $x < 1, y < 2, z < 3$, we have $a > 0, b > 0, c > 0$.

2. Express $x, y, z$ in terms of $a, b, c$: $x = 1 - a$ $y = 2 - b$ $z = 3 - c$

3. Substitute into the expression: The expression becomes $P = a \cdot b \cdot c \cdot ( (1 - a) + \frac{2 - b}{2} + \frac{3 - c}{3} )$. Simplify the last term: $x + \frac{y}{2} + \frac{z}{3} = (1 - a) + (1 - \frac{b}{2}) + (1 - \frac{c}{3}) = 3 - a - \frac{b}{2} - \frac{c}{3}$. Let $d = 3 - a - \frac{b}{2} - \frac{c}{3}$. The constraint $x + \frac{y}{2} + \frac{z}{3} > 0$ means $d > 0$. So, the expression to maximize is $P = a \cdot b \cdot c \cdot d$.

4. Apply AM-GM inequality: To apply AM-GM effectively, we need a product of terms whose sum is constant. Consider the following four positive terms: $u_1 = a$ $u_2 = \frac{b}{2}$ $u_3 = \frac{c}{3}$ $u_4 = 3 - a - \frac{b}{2} - \frac{c}{3}$

The sum of these terms is: $u_1 + u_2 + u_3 + u_4 = a + \frac{b}{2} + \frac{c}{3} + (3 - a - \frac{b}{2} - \frac{c}{3}) = 3$. This sum is a constant.

Now, express $P$ in terms of $u_1, u_2, u_3, u_4$: $a = u_1$ $b = 2u_2$ $c = 3u_3$ $P = u_1 \cdot (2u_2) \cdot (3u_3) \cdot u_4 = 6 \cdot u_1 u_2 u_3 u_4$.

By the AM-GM inequality, for non-negative numbers $u_1, u_2, u_3, u_4$: $\frac{u_1 + u_2 + u_3 + u_4}{4} \ge \sqrt[4]{u_1 u_2 u_3 u_4}$ Substituting the sum: $\frac{3}{4} \ge \sqrt[4]{u_1 u_2 u_3 u_4}$ Raising both sides to the power of 4: $\left(\frac{3}{4}\right)^4 \ge u_1 u_2 u_3 u_4$ $u_1 u_2 u_3 u_4 \le \frac{81}{256}$.

The maximum value of $u_1 u_2 u_3 u_4$ is $\frac{81}{256}$. The maximum value of $P$ is $6 \cdot \frac{81}{256} = \frac{243}{128}$.

5. Find the values of $x_0, y_0, z_0$ for which the maximum occurs: The equality in AM-GM holds when all terms are equal: $u_1 = u_2 = u_3 = u_4$. Since their sum is 3, each term must be $\frac{3}{4}$. $u_1 = a = \frac{3}{4}$ $u_2 = \frac{b}{2} = \frac{3}{4} \implies b = \frac{6}{4} = \frac{3}{2}$ $u_3 = \frac{c}{3} = \frac{3}{4} \implies c = \frac{9}{4}$ And $u_4 = 3 - a - \frac{b}{2} - \frac{c}{3} = 3 - \frac{3}{4} - \frac{3}{4} - \frac{3}{4} = 3 - \frac{9}{4} = \frac{3}{4}$, which is consistent.

Now, find $x_0, y_0, z_0$: $x_0 = 1 - a = 1 - \frac{3}{4} = \frac{1}{4}$ $y_0 = 2 - b = 2 - \frac{3}{2} = \frac{1}{2}$ $z_0 = 3 - c = 3 - \frac{9}{4} = \frac{3}{4}$

6. Calculate the required expression: We need to find the value of $\frac{z_0}{x_0y_0}$. $\frac{z_0}{x_0y_0} = \frac{3/4}{(1/4)(1/2)} = \frac{3/4}{1/8}$ $\frac{z_0}{x_0y_0} = \frac{3}{4} \times 8 = 3 \times 2 = 6$.

The final answer is $\boxed{6}$.