A linear inequality compares a changing total with a limit: a fixed fee is paid once, while the per-item cost grows with the item count. Set the cost equal to the budget in a Desmos regression to find the cutoff, then choose the greatest whole-number count the budget allows. Rounding the cutoff up can put the purchase over budget.
Hints
- Hint 1
If is the number of boxes, is the cost that grows with the count. The one-time fee is added once. What inequality keeps their total at or below the budget?
- Hint 2
The boundary is where spending exactly meets the budget. A Desmos regression can find the count at that point, even if it's a decimal. Since boxes are whole items, which nearby whole count stays under it?
Step-by-step
Find the budget cutoff
Step 1Model the spending limit
Let be the number of boxes. At $3.90 per box, the boxes cost . Add the one-time $22 fee once, not once per box. “Without exceeding” means the total can equal $180 but cannot go above it, so the inequality, a comparison allowing a range of counts, is .
- Step 2
Find the budget boundary
At the boundary, the cost equals the budget. Type in Desmos. Use because Desmos graphs but solves for ; asks it to find where the sides are equal. Under PARAMETERS, Desmos shows . More boxes cost more, so the budget allows counts at or below this cutoff.
- Step 3
Choose the greatest whole count
Boxes come in whole numbers. Since , is allowed but is too many. For a greatest whole-number count under a cutoff, take the whole number below it rather than rounding up. The club can buy at most boxes. Grid in 40.