Question 625·Medium·Algebra

Let $x$ be the number of small boxes and $y$ the number of large boxes. At least $12$ boxes means: $x+y\ge12$. Because these are box counts, $x$ and $y$ must be nonnegative whole numbers.

The small boxes contribute $18x$ cookies, and the large boxes contribute $30y$. At most $300$ cookies means: $18x+30y\le300$. Equality is allowed in both limits.

Find how many more cookies a large box holds: $30-18=12$. Count $18$ cookies for every box, then add $12$ for each large box: $18x+30y=18(x+y)+12y$. So replacing a small box with a large one adds $12$ cookies without changing the box count.

Start with the minimum cookie count for all the boxes, then add the extra cookies for each large box. Since $x+y\ge12$, the rewritten cookie total must satisfy: $18(x+y)+12y\ge18(12)+12y$. Multiply: $18(x+y)+12y\ge216+12y$. The actual total is at most $300$, so any possible order must satisfy: $216+12y\le300$. This bound rules out larger counts; you'll still check that a count at the edge works.

Type the four inequalities below into Desmos. Their overlap, the region satisfying every limit, reaches its highest point at the intersection $(5,7)$; click it to see the coordinates. That's $5$ small and $7$ large boxes, or $12$ boxes total. The point lies in both shaded regions, so this order meets both limits. The bound above increases with $y$, so no larger count can work either. The maximum is $7$ large boxes. Choice C.