Question 520·Medium·Algebra

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A local print shop charges a one-time setup fee of $28 and $1.35 per flyer. Maya has a budget of $140.

What is the greatest number of flyers she can have printed without exceeding her budget?

Your answer

A linear cost model adds a one-time fee to a per-item charge. “Without exceeding” makes the budget an upper limit, so use a Desmos regression to find the count that would spend exactly the budget. That count may be a decimal, but items come in whole numbers. Choose the greatest whole number below the limit, not the nearest rounded number.

Hints

Hint 1
A fixed fee is charged once, while a price per flyer repeats. If $n$ is the number of flyers, which charge gets multiplied by $n$ ?
Hint 2
The budget boundary is the count that would spend exactly the available money. Find it with a Desmos regression, then ask which whole-number count stays on the affordable side.

Step-by-step

Find the budget cutoff
Step 1
Write the cost limit
Let $n$ be the number of flyers. At $1.35 per flyer, the flyers cost $1.35n$ . The $28 setup fee is paid once, so add $28$ , not $28n$ . “Without exceeding” means the total can be at most $140:
$28+1.35n\le 140.$
Step 2
Find where the budget runs out
The boundary is the count that would use exactly the budget:
$28+1.35n=140.$
Type $28+1.35n_1\sim140$ in Desmos. The subscript lets Desmos fit $n_1$ , and $\sim$ tells it to match the two sides. Under PARAMETERS, it shows $n_1\approx82.963$ . Each extra flyer raises the cost, so counts above this boundary exceed the budget.
Step 3
Check the whole numbers beside the cutoff
Flyers must come in whole numbers, so check $82$ and $83$ . Type $28+1.35(82)$ and $28+1.35(83)$ ; Desmos shows $138.7$ and $140.05$ . The first bill fits the $140 budget, but the next is over it. Maya’s greatest possible number is $82$ flyers. Grid in 82.

Question 520·Medium·Algebra

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