Using the inequality for weight, $4v + 5m \le 900$, and knowing that $v \ge 0$, think about how large $m$ can be, even in the most extreme case where you minimize or eliminate vegetarian boxes.

Once you find an upper bound on $m$ from the weight constraint, try choosing $v = 0$ (all meat boxes) with that value of $m$ and check whether the budget and the minimum of $150$ total boxes are still satisfied.

Make sure you are choosing values of $v$ and $m$ that increase the number of meat boxes, even if that means having fewer vegetarian boxes, as long as all inequalities are satisfied.

In an expression line, type 900/5 and look at the numerical result. This is how many meat boxes you could have if all boxes were meat and you used exactly the $900$ pounds of weight.

In a second expression line, type 3200/17 and compare this value with the result from 900/5. The smaller of these two values is the stricter upper limit on how many meat boxes you can afford while either all boxes are meat or the budget is the only constraint.

Use the smaller value from Step 2 as a candidate for the number of meat boxes. In a new expression line, multiply it by 17 to check that the cost is at most 3200, and compare it to 150 to verify it meets the minimum total boxes requirement. The candidate value that satisfies all these checks is the maximum possible number of meat boxes.

Let $v$ be the number of vegetarian boxes and $m$ be the number of meat boxes.

Use the information to write inequalities:

• Budget: each vegetarian box costs $12$ dollars and each meat box costs $17$ dollars, with a total budget of $3200$ dollars:

• Weight: each vegetarian box weighs $4$ pounds and each meat box weighs $5$ pounds, with a maximum of $900$ pounds:

• Minimum total boxes: at least $150$ boxes in total:

• Also, $v \ge 0$ and $m \ge 0$ (you cannot make a negative number of boxes).

We want to maximize $m$ while satisfying all these inequalities.

The key inequality that directly limits how many meat boxes you can have is the weight constraint:

Because $v \ge 0$, we know $4v \ge 0$. That means $4v + 5m$ is at least $5m$:

So it must be true that

Divide both sides by $5$ to get an upper bound on $m$:

So, no matter how many vegetarian boxes are made, the number of meat boxes $m$ cannot be larger than $900/5$.

Now we need to see whether we can actually reach $m = \dfrac{900}{5}$ without breaking the other conditions.

The easiest way to maximize meat boxes is to make no vegetarian boxes, so let $v = 0$.

• Weight with $v = 0$:

Setting $5m = 900$ gives $m = \dfrac{900}{5}$, which exactly uses the full weight limit.

• Budget with $v = 0$ and $m = \dfrac{900}{5}$:

This is within the $3200$ dollar budget, since $3060 \le 3200$.

• Minimum total boxes with $v = 0$ and $m = \dfrac{900}{5}$:

Check that this is at least $150$:

So the minimum of $150$ boxes is satisfied.

Thus $m = \dfrac{900}{5}$ meat boxes is feasible: it meets the weight limit, the budget, and the minimum number of boxes.

From Step 2, $m$ cannot be greater than $\dfrac{900}{5}$, and from Step 3 we found that $m = \dfrac{900}{5}$ is actually possible while satisfying all constraints.

Compute this value:

Therefore, the maximum number of meat boxes the charity can prepare is $180$.