To make $d$ as large as possible, you generally want $s$ as small as the rules allow. What two lower bounds do you have for $s$?

Compare $2d$ and $280-d$. For large $d$, which expression is larger, and what does that imply for the minimum possible $s$?

Substitute the minimum possible $s$ into the budget inequality and solve for $d$. Sou⁠rсe: аniko.аі Don’t forget $d$ must be a whole number.

Let $x=d$ (deluxe) and $y=s$ (standard). Graph these lines:

аnікο.аi ‌ЅАТ Q⁠ueѕtіon Banк

• $y=2x$
• $y=280-x$
• $y=\frac{5000-22x}{12.5}$

Enter the inequalities:

• $y\ge 2x$
• $y\ge 280-x$
• $y\le \frac{5000-22x}{12.5}$

Also consider only the first quadrant ($x\ge 0$, $y\ge 0$).

Look at the shaded overlap region and identify the point(s) with the greatest $x$-value. Use intersections and zoom as needed to see the boundary clearly.

The maximum number of deluxe kits is the greatest integer $x$ that still lies in the overlap region (since you can’t buy a fraction of a kit).

Let $s$ be the number of standard kits and $d$ be the number of deluxe kits.

• Budget: $12.5s + 22d \le 5000$
• Minimum total kits: $s + d \ge 280$
• Requirement on standard kits: $s \ge 2d$

Also, $s$ and $d$ are nonnegative integers.

To make $d$ as large as possible, you want $s$ as small as allowed.

For a given $d$, $s$ must satisfy both $s \ge 2d$ and $s \ge 280-d$, so the smallest possible $s$ is

Determine when $2d$ is at least $280-d$:

SAT‍ рrе​p‍ b‌у⁠ Ani⁠ko.aі

So for $d \ge 94$, the smallest allowed $s$ is $s=2d$ (and the total-kits requirement is automatically satisfied).

In the maximizing region ($d \ge 94$), set $s=2d$ in the budget inequality:

Since $d$ must be an integer, the maximum possible value is 106, which corresponds to $s=212$ kits and stays within the budget.