A wholesaler who deals in household chemicals and paper products has a container with a volume of $12 \mathrm{~m}^{3}$ and can accommodate 5 tons of goods. The chemical products occupy $1 \mathrm{~m}^{3}$ per ton, while the paper products occupy $3 \mathrm{~m}^{3}$ per ton. The profit is one hundred thousand forints per ton of chemical products, and two hundred thousand forints per ton of paper products.

What is the maximum profit the wholesaler can make from selling one container of goods?
I. solution. If we take $v$ tons of chemical waste and $p$ tons of paper waste, then the truck's capacity can be described by the following conditions:

$$
\begin{aligned}
v+p & \leq 5 \\
v+3 p & \leq 12
\end{aligned} \quad(v, p \geq 0)
$$

If $H$ represents the profit, then

$$
\frac{H}{10^{5}}=v+2 p
$$

$H$ is a strictly increasing function of both $v$ and $p$; therefore, if $v_{1} \geq v_{0}$ and $p_{1} \geq p_{0}$, and at least one of these inequalities is strict, then the profit for $v_{1}, p_{1}$ is greater than for $v_{0}, p_{0}$. Thus, it is optimal to increase the values of $v$ and $p$ until we get equality in (1) or (2): $H$ can only be maximal for such values. Accordingly, we can distinguish two cases.

1. case:

$$
\begin{gathered}
v+p=5 \\
v+3 p \leq 12
\end{gathered}
$$

Express $v$ from (1) and substitute it into (2):

$$
\begin{aligned}
(5-p)+3 p & \leq 12 \\
p & \leq 3.5
\end{aligned}
$$

Thus,

$$
\frac{H}{10^{5}}=(5-p)+2 p=p+5 \leq 8.5
$$

So in this case, the maximum of $H$ is $8.5 \cdot 10^{5}=850000$, which occurs when $p=3.5$ and $v=5-p=1.5$.

2. case:

$$
\begin{gathered}
v+p \leq 5 \\
v+3 p=12
\end{gathered}
$$

Proceeding similarly to the previous case, from (2) we get $v=12-3 p$, substituting this into (1):

$$
\begin{gathered}
\quad(12-3 p)+p \leq 5, \quad p \geq 3.5, \quad \text { thus } \\
\frac{H}{10^{5}}=(12-3 p)+2 p=12-p \leq 8.5
\end{gathered}
$$

Here, equality holds precisely when $p=3.5$ and $v=12-3 p=1.5$, which coincides with the extreme value point found in the first case. Therefore, the profit is at most 850000 forints, and it is exactly this amount when $p=3.5$ and $v=1.5$.

II. solution. Plot the functions (1) and (2) in the $p, v$ coordinate system.

![](https://cdn.mathpix.com/cropped/2024_05_02_5b4618fd72057c8d3997g-1.jpg?height=800&width=539&top_left_y=1904&top_left_x=782)

To find the maximum of the profit function $v+2 p$ in units of 100,000 forints.

The coordinates of the intersection point of the two lines are $\left(\frac{7}{2} ; \frac{3}{2}\right)$. The values that satisfy the conditions fall within and on the boundary of the quadrilateral defined by the points $(0 ; 5)$, $(0 ; 0)$, $(4 ; 0)$, and $\left(\frac{7}{2} ; \frac{3}{2}\right)$. The linear function representing the profit intersects the $v$-axis at a point $C$. Different profit values correspond to different parallel lines. The maximum profit is achieved when the line passes through the point $\left(\frac{7}{2} ; \frac{3}{2}\right)$, i.e., when 3.5 tons of chemical waste and 1.5 tons of paper waste are transported in the container.