What is the smallest and what is the largest value of the expression $K=5 x-6 y+7 z$ if $x, y$ and $z$ are non-negative numbers that satisfy

$$
4 x+y+2 z=4
$$

and

$$
3 x+6 y-2 z=6 ?
$$
I. solution. Express the $K$ expression depending on three parameters in terms of a single parameter, for example, $x$. From the conditional equations, by adding and rearranging,

$$
y=\frac{10}{7}-x
$$

and using this, for example, from (2),

$$
z=\frac{9}{7}-\frac{3}{2} x
$$

Based on the condition, $x$ cannot be negative, but it can take the value 0, because when $x=0$, according to (4) and (5), $y$ and $z$ are positive, so the minimum value of $x$ is 0. On the other hand, from the fact that $y$ and $z$ are non-negative, (4) and (5) imply

$$
x \leq \frac{10}{7}, \quad \text { and } \quad x \leq \frac{6}{7}
$$

so the maximum value of $x$ is $\frac{6}{7}$.

Using (4) and (5),

$$
K=5 x-6 y+7 z=\frac{x}{2}+\frac{3}{7}
$$

the value of this expression increases with $x$, so its minimum and maximum values, by substituting the minimum and maximum values of $x$,

$$
K_{\min }=\frac{3}{7}, \quad \text { and } \quad K_{\max }=\frac{6}{7}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_faa70d35a6a3b603ae62g-1.jpg?height=540&width=680&top_left_y=1069&top_left_x=706)

II. solution. Visualize the three-variable expressions in the problem using a three-dimensional Cartesian coordinate system. (This solution assumes knowledge of certain elements of three-dimensional coordinate geometry.)

The equations (2) and (3) each define a plane in space (Figure 6). The triples of numbers that satisfy both equations are the coordinates of points lying on the intersection line of these two planes. Due to the additional conditions $x \geq 0, y \geq 0, z \geq 0$, only the part of this intersection line that lies in the first octant (and on its boundary) is relevant, which extends from the point $P_{1}(0,10 / 7,9 / 7)$ to the point $P_{2}(6 / 7,4 / 7,0)$.

The equation $5 x-6 y+7 z=K$ defines a plane for each given value of $K$, which intersects the $X$, $Y$, and $Z$ axes at

$$
a=K / 5, \quad b=-K / 6, \quad c=K / 7
$$

These intercepts are proportional to $K$, so the intersection lines of the planes corresponding to different $K$ values with the coordinate planes are parallel, and thus the planes themselves are parallel.

![](https://cdn.mathpix.com/cropped/2024_05_02_faa70d35a6a3b603ae62g-1.jpg?height=572&width=593&top_left_y=2132&top_left_x=690)

Let's represent the plane of the parallel family passing through $P_{1}$ and $P_{2}$ (Figure 7). The value of $K$ at $P_{1}$ is $3 / 7$, and at $P_{2}$ it is $6 / 7$, which are different, so different planes of the family pass through $P_{1}$ and $P_{2}$. The intercepts are

$$
\begin{array}{llll}
a_{1}=3 / 35, & b_{1}=-3 / 42, & c_{1}=3 / 49, & \text { and } \\
a_{2}=6 / 35, & b_{2}=-6 / 42, & c_{2}=6 / 49 &
\end{array}
$$

For each internal point of the segment $P_{1} P_{2}$, exactly one plane of the parallel family passes through it, lying between the two drawn planes. From the intercepts, it can be determined that $K$ increases from $P_{1}$ to $P_{2}$, so the above $K_{1}$ is the minimum, and $K_{2}$ is the maximum value.