12.44 In an arithmetic progression, the sixth term is equal to 3, and the difference of the progression is greater than 0.5. For what value of the difference of this progression is the product of the first, fourth, and fifth terms the greatest?
12.44 According to the condition, $a_{6}=a_{1}+5 d=3$, from which $a_{1}=3-5 d$. Let the product $a_{1} a_{4} a_{5}$ be denoted by $y$. Then we get $y=a_{1}\left(a_{1}+3 d\right)\left(a_{1}+4 d\right)=-10 d^{3}+51 d^{2}-72 d+27$.

To find the value of $d$ at which the function $y$ takes its maximum value, we first find the derivative

$$
y^{\prime}=-30 d^{2}+102 d-72=-6\left(5 d^{2}-17 d+12\right)
$$

and then, by solving the equation $5 d^{2}-17 d+12=0$, we find its roots $d_{1}=1 ; d_{2}=2.4$. Since by the condition $d>0.5$, we investigate the behavior of the function $y$ on the interval $(0.5 ; \infty)$. We construct the table

| Interval of change of $d$ | $(0.5 ; 1)$ | 1 | $(1 ; 2.4)$ | 2.4 | $(2.4 ; \infty)$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $y^{\prime}$ | - | 0 | + | 0 | - |
| $y$ | $\searrow$ | $\min$ | $\nearrow$ | $\max$ | $\searrow$ |

On the interval $(0.5 ; \infty)$, there is only one point of maximum of the function $y$, namely $d=2.4$. This means that on the interval $(0.5 ; \infty)$, the function $y$ reaches its maximum value at $d=2.4$.

Answer: at $d=2.4$.