Let's break down the problem step-by-step and solve it using algebraic methods, ensuring we don't refer to any Python code:
1. Identify the variables and given values:
- Charge per print (price per print): [tex]\( \$0.30 \)[/tex]
- Cost per print: [tex]\( \$0.08 \)[/tex]
- Overhead cost daily: [tex]\( \$45.00 \)[/tex]
- Desired profit per day: [tex]\( \$50.00 \)[/tex]
2. Calculate the profit per print:
The profit per print is the difference between the charge per print and the cost per print.
[tex]\[
\text{Profit per print} = \$0.30 - \$0.08 = \$0.22
\][/tex]
3. Calculate the total amount of money needed to cover the overhead costs and desired profit:
[tex]\[
\text{Total amount needed} = \text{Overhead cost} + \text{Desired profit} = \$45.00 + \$50.00 = \$95.00
\][/tex]
4. Determine the number of prints required to cover the total amount needed:
Let [tex]\( n \)[/tex] be the number of prints needed.
The total profit needed from [tex]\( n \)[/tex] prints is:
[tex]\[
n \times \text{Profit per print} \geq \text{Total amount needed}
\][/tex]
Substituting the values we have:
[tex]\[
n \times 0.22 \geq 95
\][/tex]
5. Solve for [tex]\( n \)[/tex]:
[tex]\[
n \geq \frac{95}{0.22} \approx 431.82
\][/tex]
Since the number of prints must be a whole number, we round up to the nearest whole number:
[tex]\[
n \geq 432
\][/tex]
Therefore, the kiosk must produce at least 432 prints each day to realize a profit of at least $50.00 per day.
