Question 435·Hard·Algebra

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A coffee-roasting company will buy only two types of coffee beans for a new blend: a standard bean that costs $4.80 per pound and a premium bean that costs $7.60 per pound. The shipping charge for the entire order is a flat $120. The company’s budget for the purchase, including shipping, is $2,000.

The beans are purchased in whole-pound amounts. The order must include at least 60 pounds of premium beans but no more than half of the total number of pounds purchased can be premium beans.

What is the greatest number of pounds of premium beans the company can purchase without exceeding the budget?

A system of inequalities appears when a purchase has both a budget and a required mix of two items. Use the mix rule to find the fewest cheaper items allowed for each costlier item, then put that amount into the budget. Solve the resulting cutoff in Desmos and check whole-number amounts. Don't omit a flat shipping charge or count it more than once.

Hints

Hint 1
The phrase no more than half compares premium pounds with all the pounds purchased. If $x$ is premium and $y$ is standard, what must be true about $y$ so premium doesn't make up more than half?
Hint 2
A flat fee is paid once for the entire order, regardless of its size. For a chosen number of premium pounds, use the fewest standard pounds the mix rule allows. That leaves the most room in the budget.
Hint 3
A ceiling is the largest amount before a limit is broken. Find the premium-pound ceiling using the cheapest allowed mix, then check whole-pound amounts on either side of it.

Step-by-step

Use the cheapest allowed mix
Step 1
Translate the half rule
Let $x$ be premium pounds and $y$ be standard pounds. If premium were more than standard, premium would make up more than half the order. So no more than half premium means $y\ge x$ .
Step 2
Write the budget limit
The beans cost $4.80y+7.60x$ dollars, and the flat shipping charge adds $120$ dollars once. Since the total can't exceed the budget, the budget inequality is $4.80y+7.60x+120\le2000$ .
Step 3
Use the fewest standard pounds allowed
For any fixed $x$ , more standard pounds cost more money. The cheapest valid order has $y=x$ ; if even that order is over budget, no order with that many premium pounds can work. Substitute $y=x$ into the budget: $4.80x+7.60x+120\le2000$ . To maximize one count, minimize the other as far as the rules allow.
Step 4
Find the budget cutoff
Type $4.80x_1+7.60x_1+120\sim2000$ in Desmos. The $\sim$ asks Desmos to find where the cost equals the budget, and $x_1$ is the unknown it fits. Under PARAMETERS, Desmos shows $x_1\approx151.613$ . That's the cutoff: the cost rises as premium pounds increase, so a larger amount can't fit.
Step 5
Check the neighboring whole-pound amounts
The beans come in whole-pound amounts. Type $(4.80+7.60)\cdot151+120$ and $(4.80+7.60)\cdot152+120$ ; Desmos prints $1992.40$ and $2004.80$ . These use the cheapest mix, with equal pounds of each bean. The first fits the $2000$ budget and meets the minimum of $60$ premium pounds. The next already exceeds the budget, so the greatest amount is $151$ pounds of premium beans. Choice B.

Question 435·Hard·Algebra

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