Write three inequalities: one for the total number of boxes, one for the total cost using $7$ and $4$, and one for the total calories using $400$ and $250$. Make sure you use the correct inequality symbols (at least, at most).

If your goal is to maximize the number of premium boxes, does adding extra standard boxes (beyond the minimum of 200 total boxes) ever help with staying under budget or under the calorie limit? Use this idea to write $s$ in terms of $p$ and substitute into the other inequalities.

After substituting, you should get inequalities that bound $p$ from above. Find the largest integer that satisfies all those bounds, and then quickly check that the next integer up breaks at least one constraint.

In Desmos, let $x$ be the number of premium boxes and $y$ the number of standard boxes. Enter these inequalities:

• x + y >= 200
• 7x + 4y <= 1000
• 400x + 250y <= 65000
• x >= 0
• y >= 0 This will shade the feasible region where all constraints are met.

Graph the boundary lines x + y = 200 and 7x + 4y = 1000 (you can enter them as equalities). Tap their intersection point; note the $x$-coordinate of this point, because it represents where the minimum-total-boxes line meets the budget limit.

Because you want to maximize the number of premium boxes $x$, look for the largest integer less than or equal to the $x$-coordinate from the intersection in Step 2, and verify that this integer (paired with $y = 200 - x$) still lies in the shaded feasible region. That integer value of $x$ is the answer.

Let $p$ be the number of premium snack boxes and $s$ be the number of standard snack boxes.

Translate each condition:

• At least 200 boxes:
  $p + s \ge 200$.
• Budget is at most $1000:
  $7p + 4s \le 1000$.
• Calories at most $65000$:
  $400p + 250s \le 65000$.
• Also, $p \ge 0$ and $s \ge 0$.

We want to maximize $p$ subject to these inequalities.

To maximize $p$, it never helps to add extra standard boxes, because each standard box costs money and adds calories.

So it is best to have exactly 200 total boxes (the minimum allowed):

Solve for $s$:

Now we can substitute this expression for $s$ into the other inequalities so everything is in terms of $p$.

Substitute $s = 200 - p$ into the budget inequality $7p + 4s \le 1000$:

Simplify:

So the budget constraint becomes

This gives an upper bound on how large $p$ can be.

Now use $s = 200 - p$ in the calorie inequality $400p + 250s \le 65000$:

Simplify:

The calorie constraint allows $p \le 100$, but the budget is stricter: $p \le \frac{200}{3}$.

The largest integer less than or equal to $\frac{200}{3}$ is 66. With $p = 66$, we have $s = 200 - 66 = 134$, and you can check:

• Budget: $7(66) + 4(134) = 998 \le 1000$.
• Calories: $400(66) + 250(134) = 59900 \le 65000$.

If you try $p = 67$, even with $s = 133$ (the minimum), the cost exceeds the $1000 budget. Therefore, the greatest possible number of premium boxes is 66, which corresponds to answer choice B.