How to find the least number of days by month for a machine to reduce costs?

The problem is as follows:

A baker has to use each day an electric mixer to cook his famous strawberry muffins. He has two options as indicated in the table from below:

$$\begin{array}{lll} \textrm{Brand}&\textrm{Monthly rent}&\textrm{Operation cost and daily maintenance}\\ \textrm{Axial tech}&\textrm{900 USD}&\textrm{55 USD}\\ \textrm{Boost tech}&\textrm{800 USD}&\textrm{60 USD}\\ \end{array}$$

What is the least number of days in a month the baker can use Axial brand so that it result cheaper than using Boost brand?.

The existing alternatives given are:

$$\begin{array}{ll} 1.&19\\ 2.&20\\ 3.&21\\ 4.&22\\ \end{array}$$

For this specific problem I'm confused on how to use the figures indicated in the table. In my attempt what I tired to do was to find the total cost for each brand.

For Axial:

$$Cost = 900 + 55 \times 30 = 2550$$

For Boost:

$$Cost = 800 + 60 \times 30 = 2600$$

But as it stands it looks that for a month (assuming $$30$$ days) Axial brand is cheaper.

Then this discrepancy made me confused. Can somebody help me with the right interpretation for this problem?.

Let $$n$$ be the minimal number of days in the month which is required. Also, assume you're paying for a full month's rent. Then you want to find the smallest integral $$n$$ where
\begin{aligned} 900 + 55n & \lt 800 + 60n \\ 100 & \lt 5n \\ 20 & \lt n \end{aligned}
The smallest such $$n$$ is $$21$$, i.e., answer $$3$$.
• In general, when a question asks for "the number of days", it's assumed they're referring to an integral #, so $21$ is smallest. Also, more specifically here, the table's 2nd column is "Operation cost and daily maintenance". As this is specified on a per day basis, it's reasonable to assume you can't split it into smaller amounts, e.g., half a day. This is partially due to things the assumption the baker uses the electric mixer for only part of a day, but the maintenance costs only occur while in use, so if you count part of a day, would it be while it's in use, not in use or a mixture? Sep 13, 2019 at 23:01
• @ChrisSteinbeckBell Also, since the question asks for "the least number of days in a month the baker can use Axial brand so that it result cheaper than using Boost brand", you only have to find the first integral number where this is true. Although you are correct that the overall difference increases to infinity, since there's a net saving every month, this is not required for answering the question, since the result would still be $21$ even if, for example, the later monthly costs were changed so the Axial brand became more expensive again after $25$ days. Sep 13, 2019 at 23:13