The price of an incandescent lamp is $8 \mathrm{Ft}$, its lifetime as a function of voltage is $E=2 \cdot 10^{6} \cdot 2^{-U /(10 \mathrm{~V})}$ hours. Its efficiency (candela/watt) depends on the voltage: $G=4 \cdot 10^{-5}(U / 1 \mathrm{~V})^{1.7}$ candela/watt. The price of $1 \mathrm{kWh}$ of energy is $2.1 \mathrm{Ft}$. At what voltage should we operate the lamp to get the cheapest light over a longer period? The resistance of the filament is $R=200 \Omega$; assume it is independent of the voltage.
If the resistance of the light bulb is independent of the voltage, its power can be given by $U^{2} / R$. The operating cost of the light bulb over its entire lifetime is:

$$
E \cdot U^{2} /\left(R \cdot 10^{3}\right) \cdot a
$$

where $E\left(=E_{0} \cdot 2^{-0.1 U}\right)$ is the lifespan of the light bulb, $a(=2.1 \text{ Ft} / \text{kWh})$ is the operating cost, and $R(=200 \Omega)$ is the resistance of the light bulb.

![](https://cdn.mathpix.com/cropped/2024_05_02_c3596721a2838b4bba0cg-1.jpg?height=940&width=903&top_left_y=381&top_left_x=600)

The light energy emitted per unit time is

$$
U^{2} /\left(R \cdot 10^{3}\right) G
$$

where $G\left(=G_{0} U^{1.7}\right)$ is the efficiency of light emission.

We can operate the light bulb at the lowest cost if the cost per unit time for unit light emission is the smallest, i.e., if the function

$$
f(U)=\frac{E U^{2} /\left(R \cdot 10^{3}\right) \cdot a + A}{E G U^{2} /\left(R \cdot 10^{3}\right)}
$$

has a minimum (where $A$ is the price of the light bulb). Substituting the expressions and simplifying, we get:

$$
f(U)=\alpha U^{-1.7}\left(1+\beta 2^{0.1 U} U^{-2}\right)
$$

where $\alpha=42.5$ and $\beta=0.372$, which we obtained based on the following data: $G_{0}=4 \cdot 10^{-2}$ candela $/ \text{kW}$, $a=2.1$ $\text{Ft} / \text{kWh}$, $E_{0}=2.048 \cdot 10^{6} \text{ h}$, $R=200 \Omega$, $A=8 \text{ Ft}$.

Our task is to find the minimum of the $f(U)$ function with respect to $U$. The simplest way to do this is to tabulate the $f(U)$ function. The part of the $f(U)$ function near its minimum is shown in the figure. From this, we can determine the $U$ value corresponding to the minimum, which is $\approx 138.5 \text{ V}$. Of course, we can also find the minimum by differentiation. In this case, we need to solve the equation $\frac{d f(U)}{d U}=0$. It is useful to use the transformation

$$
2^{0.1 U}=e^{0.1 U \ln 2}
$$

Using the power and chain rule for differentiation, we get:

$$
f^{\prime}(U)=-\frac{1.7 U^{-2.7} a}{G_{0}}+\frac{A K}{E_{0} G_{0} 10^{3}}\left(0.1 \cdot \ln 2 \cdot e^{0.1 U \ln 2} \cdot U^{-3.7}-e^{0.1 U \ln 2} \cdot 3.7 U^{-4.7}\right)
$$

The condition for the minimum is that $f^{\prime}(U)=0$; after simplifications, we get:

$$
e^{-0.1 U \cdot \ln 2}=\frac{A R U^{-1} 10^{3}}{1.7 \cdot a E_{0}}\left(0.1 \cdot \ln 2-3.7 U^{-1}\right)
$$

The solution to this equation can only be found by trial and error. A relatively simple way is to tabulate both sides and find the $U$ values where the two sides are approximately equal. We can see that this calculation is not simpler than plotting the $f(U)$ function.