Question 390·Hard·Algebra

Let $x$ be the number of vegetarian boxes and $y$ the number of meat boxes. The given costs make the total cost $12x+17y$ dollars, so the budget is a cap:

Each vegetarian box adds $4$ pounds and each meat box adds $5$ pounds. So the weight limit is:

Together, the charity makes $x+y$ boxes. At least $150$ means $150$ counts, as does any larger total:

The charity can't prepare a negative number of either kind of box, so add $x\ge0$ and $y\ge0$. Box counts must also be whole numbers; Desmos will shade fractional counts too.

Type all five inequalities. The feasible region is where every shading overlaps. Desmos shows its top edge meeting the weight boundary at $x=0$. That's the candidate to check, rather than a point that meets only the weight limit.

Subtract $4x$ from the weight limit:

Since $x\ge0$, the right side can be no larger than $900$:

The ceiling can be reached when $x=0$, making $5y=900$. A maximum needs both a ceiling and a combination that reaches it.

With $x=0$ and $5y=900$, the meat boxes cost $17(900/5)$ dollars. Add $17(900/5)$ in Desmos; it prints $3060$. That is within the $3,200 budget, so cost doesn't rule out this top point.

At $x=0$, the total box count is $900/5$. Add $900/5-150$ in Desmos; it prints $30$. The total is 30 boxes above the minimum, so this point meets that requirement too.

To find the count at that top point, add $5y_1\sim900$. Use $y_1$ for a number Desmos can fit and $\sim$ to ask it to solve; under PARAMETERS, it shows $y_1=180$. With $x=0$, that's a whole-number count that meets every limit, and the weight ceiling rules out more. The maximum is $180$ meat boxes. Grid in 180.