The $2{,}000 budget includes a flat $120 shipping fee. How much money is left for the beans themselves, and what inequality does that give you with $4.8s$ and $7.6p$?

Translate "no more than half of the total pounds can be premium" into an inequality involving $p$ and $s$. Then simplify it to a relationship between $p$ and $s$ only (no fractions).

Once you know that standard beans must be at least as many pounds as premium beans, what is the smallest possible value of $s$ for a given $p$? Substitute that into your cost inequality to get a single inequality in $p$, then remember that $p$ must be a whole number.

Treat the number of pounds of standard beans as $x$ and premium beans as $y$. In Desmos, enter these inequalities on separate lines:

• 4.8x + 7.6y <= 1880
• y >= 60
• y <= x This graphs the region of $(x,y)$ pairs that meet the budget and bean-mix constraints.

Also graph the boundary lines 4.8x + 7.6y = 1880 and y = x. Their intersection lies on the edge of the feasible region and represents the point where the budget and the "no more than half premium" condition are both tight. In the graph, click that intersection point and read off the $y$-coordinate; this gives the maximum possible value of $y$ before considering the whole-number requirement.

In Desmos, you can also type 1880/12.4 in a new line to see the same upper bound for $y$. Then, because the number of pounds must be a whole number, take the greatest integer that does not exceed that decimal value and check that it still satisfies $y \ge 60$ and the cost inequality.

Let

• $p$ = number of pounds of premium beans
• $s$ = number of pounds of standard beans

The total budget is $2{,}000, and shipping is a flat $120, so the maximum that can be spent on beans is

The bean cost is $4.80 per pound for standard and $7.60 per pound for premium, so the cost inequality for the beans is

The problem gives two more constraints:

1. At least 60 pounds of premium beans:

2. No more than half of the total pounds can be premium. The total pounds is $s + p$, so

Multiply both sides by $2$:

This means there must be at least as many standard pounds as premium pounds.

We want to make $p$ as large as possible while still satisfying all constraints.

From $p \le s$, for any fixed $p$, the smallest allowed $s$ is $s=p$. Since standard beans also cost money, keeping $s$ as small as possible (while still $\ge p$) gives the lowest total cost for a given $p$.

So, to find the largest possible $p$ that still fits the budget, set $s=p$ in the cost inequality:

Combine like terms:

Solve this inequality symbolically for $p$:

Now evaluate the fraction

To avoid decimals, multiply numerator and denominator by $10$:

Divide:

• $124 \cdot 150 = 18{,}600$
• $18{,}800 - 18{,}600 = 200$
• $124$ goes into $200$ once, so $124 \cdot 151 = 18{,}724$

So

which gives

Because $p$ must be a whole number of pounds, the greatest possible $p$ is $151$.

Check quickly:

• Use $p = 151$ and $s = 151$ (this keeps premium at exactly half the total):

• Total cost including shipping is $1{,}872.4 + 120 = 1{,}992.4 \le 2{,}000$.
• $p = 151$ is at least $60$ and is not more than half the total pounds.

Therefore, the greatest number of pounds of premium beans the company can purchase is 151.