From the cost inequality, think about what happens if there are no pretzels. What inequality do you get for the maximum possible number of almond packets?

Solve the protein inequality for the number of pretzel packets in terms of the number of almond packets. Then, use that expression in the cost inequality to get an inequality involving only the number of almond packets.

You should end up with an upper bound and a lower bound for the number of almond packets. Combine them to pin down the exact value of the minimum number of almond packets.

Let $x$ represent the number of almond packets and $y$ the number of pretzel packets. In Desmos, enter the inequalities:

• 0.7x + 0.45y <= 280
• 6x + 2y >= 2400
• x >= 0
• y >= 0 This will shade the region of points $(x,y)$ that satisfy all the constraints.

Zoom and pan until you clearly see the shaded region where all four inequalities overlap. In this problem, the overlap collapses to a single point on the graph, where the budget and protein constraints are both tight (turned into equalities).

Click on that single overlapping point to see its coordinates $(x,y)$. The $x$-coordinate of this point is the number of almond packets in the only combination that satisfies both the protein requirement and the budget, and thus is the minimum number of almond packets needed.

Let

• $a$ = number of almond packets
• $p$ = number of pretzel packets

Total cost is at most $280:

Total protein is at least $2400$ grams:

We also know $a \ge 0$ and $p \ge 0$ since you cannot buy a negative number of packets.

From the cost inequality

notice that $0.45p \ge 0$ because $p \ge 0$. That means

Dividing both sides by $0.70$ gives

So there is a maximum possible number of almond packets; it cannot exceed this value.

From the protein inequality

solve for $p$:

For any fixed $a$, the smallest number of pretzel packets that still meets the protein requirement is $p = 1200 - 3a$ (as long as this is nonnegative). Using more pretzels would only increase the total cost.

So, to see whether a given $a$ can fit in the budget at all, plug this minimum $p$ into the cost inequality:

Simplify:

This gives a lower bound on $a$.

From the previous steps we have two bounds on $a$:

• From the budget alone: $a \le \dfrac{280}{0.70}$.
• From combining budget and protein: $0.65a \ge 260$.

Now solve the inequality from Step 3:

From Step 2, $a \le \dfrac{280}{0.70} = 400$. Together these force

With $a = 400$, the protein requirement is met by almonds alone: $6 \cdot 400 = 2400$ grams, and the cost is exactly $0.70 \cdot 400 = 280$ dollars, leaving no money for pretzels (so $p = 0$).

Therefore, the minimum number of almond packets the nutritionist must purchase is 400.