Question

1. the drama club is running a lemonade stand to raise money for its new production. a local grocery store donated cans of lemonade and bottles of water. cans of lemonade sell for $2 each and bottles of water sell for $1.50 each. the club needs to raise at least $500 to cover the cost of renting costumes. the students can accept a maximum of 360 cans and bottles.

part a. let x represents the number of cans of lemonade and y represents the number of bottles of water. write a system of inequalities that can be used to represent this situation.

part b. the club sells 144 cans of lemonade. what is the least number of bottles of water that must be sold to cover the cost of renting costumes? justify your answer.

Explanation:

Step1: Define total item limit

The total number of cans and bottles cannot exceed 360.
$x + y \leq 360$

Step2: Define fundraising minimum

Total revenue must be at least $500.
$2x + 1.5y \geq 500$

Step3: Define non-negative constraints

Quantities cannot be negative.
$x \geq 0, \, y \geq 0$

Step4: Substitute x=144 for Part B

Plug $x=144$ into revenue inequality.
$2(144) + 1.5y \geq 500$

Step5: Calculate fixed revenue

Compute revenue from lemonade cans.
$288 + 1.5y \geq 500$

Step6: Isolate y term

Subtract 288 from both sides.
$1.5y \geq 500 - 288$
$1.5y \geq 212$

Step7: Solve for y

Divide both sides by 1.5, round up.
$y \geq \frac{212}{1.5} \approx 141.33$
Since partial bottles can't be sold, $y=142$.

Answer:

Part A:

The system of inequalities is:
$x + y \leq 360$
$2x + 1.5y \geq 500$
$x \geq 0, \, y \geq 0$

Part B:

The least number of bottles of water that must be sold is 142. When 144 cans are sold, they generate $288 in revenue. To reach the $500 minimum, $212 more is needed. Dividing $212 by $1.50 per bottle gives approximately 141.33, so 142 full bottles are required to meet or exceed the fundraising goal.