A plant increases its annual production by $6 \%$ in the first year of the five-year plan, and by an additional $8 \%$ in the second year. Over the next three years, they want to increase production by the same percentage each year, so that the overall increase in production over the course of the plan averages $10 \%$ annually. By what percentage should they increase production each year?
Let the production in the year before the plan be $T$, then in the first year of the plan, the production was $T_{1}=T+\frac{6}{100} T=T \cdot 1.06$. In the second year of the plan, $T_{1}$ increased by $8\%$, so the production in the second year of the plan was $T_{2}=T_{1} \cdot 1.08=T \cdot 1.06 \cdot 1.08$. This $T_{2}$ production is to be increased over 3 years, annually by a constant $x\%$, so that in the final year of the plan, the production $T_{5}=T_{2}\left(1+\frac{x}{100}\right)^{3}$ equals $T \cdot 1.1^{5}$. Therefore,

If we use $q$ instead of $1+\frac{x}{100}$ for simplicity, then $T_{2} q^{3}=T \cdot 1.06 \cdot 1.08 \cdot q^{3}=T \cdot 1.1^{5}$,

$$
q^{3}=\frac{1.1^{5}}{1.06 \cdot 1.08}
$$

and thus

$$
q=\sqrt[3]{\frac{1.1^{5}}{1.06 \cdot 1.08}}=1.121
$$

and therefore

$$
x=100(q-1)=12.1\%
$$

So, the production must be increased by $12.1\%$ annually over the remaining 3 years.

Remark: The text can also be interpreted to mean that instead of the production in the fifth year, it is about the total production over five years. In this case (after simplifying with $T$, we arrive at the equation

$$
1.06+1.06 \cdot 1.08\left(1+q+q^{2}+q^{3}\right)=1.1 \frac{1.1^{5}-1}{0.1}=6.71561
$$

from which

$$
q^{3}+q^{2}+q+1=\frac{6.71561-1.06}{1.1448}=4.94
$$

We rounded the right-hand side fraction to two decimal places, considering that we will solve the equation graphically and could not expect greater accuracy.

If we multiply both sides by $(q-1)$, then

$$
q^{4}-1=4.94 q-4.94
$$

or

$$
q^{4}=4.94 q-3.94
$$

Multiplying by $(q-1)$ introduces a root of 1 to this fourth-degree equation, which can be easily solved graphically, and the other roots coincide with the roots of the original third-degree equation.

For the graphical solution, it is practical to divide both sides of our equation by 10:

$$
0.1 q^{4}=0.494 q-0.394
$$

Let's plot the functions $y_{1}=0.1 q^{4}$ and $y_{2}=0.494 q-0.394$:

| $q$ | 1 | 1.1 | 1.2 |
| ---: | ---: | :--- | :--- |
| $y_{1}$ | 0.1 | 0.146 | 0.208 |
| $x_{2}$ | 0.1 | 0.149 | 0.198 |

![](https://cdn.mathpix.com/cropped/2024_05_02_38b229127e95a9e26650g-1.jpg?height=591&width=607&top_left_y=1911&top_left_x=756)

From the graph, it can be seen that the line intersects the fourth-degree curve at a point around $q=1.1425$, in addition to the point $q=1$. (It is easy to verify with a four-digit function table that at $q=1.141$, the fourth-degree function is still below the line, and at $q=1.144$, it is definitely above the line, so $1.141<q<1.144$.)

So, in this case, the production must be increased by $14.25\%$ annually in the final year of the plan.