Question

select the correct answer. ms. waller receives orders for posy bouquets and cascade bouquets. it usually takes her 15 minutes to make one posy bouquet and 20 minutes to make one cascade bouquet. for today’s orders, each posy bouquet must have 12 stems and each cascade bouquet must have 6 stems. ms. waller has no more than 300 minutes to make the bouquets, and she can procure no more than 180 stems in all. ms. waller makes a profit of $10 on each posy bouquet and $7 on each cascade bouquet. how many posy bouquets and cascade bouquets should she deliver to maximize her profit? a. ms. waller should make 6 posy bouquets and 12 cascade bouquets to maximize her profit. b. ms. waller should make 15 posy bouquets and 0 cascade bouquets to maximize her profit. c. ms. waller should make 0 posy bouquets and 15 cascade bouquets to maximize her profit. d. ms. waller should make 12 posy bouquets and 6 cascade bouquets to maximize her profit.

Explanation:

Step1: Define variables

Let $x$ = number of posy bouquets, $y$ = number of cascade bouquets.

Step2: List constraints

Time constraint: $15x + 20y \leq 300$ (simplify to $3x + 4y \leq 60$)
Stem constraint: $12x + 6y \leq 180$ (simplify to $2x + y \leq 30$)
Non-negativity: $x \geq 0$, $y \geq 0$

Step3: Define profit function

Profit $P = 10x + 7y$

Step4: Find feasible vertices

Solve intersection of constraints:

1. $x=0,y=0$: $P=0$
2. $x=0, 3(0)+4y=60 \to y=15$: $P=10(0)+7(15)=105$
3. $y=0, 2x+0=30 \to x=15$: $P=10(15)+7(0)=150$
4. Solve

$$\begin{cases}3x+4y=60\\2x+y=30\end{cases}$$

From second equation: $y=30-2x$. Substitute into first:
$3x + 4(30-2x)=60 \to 3x+120-8x=60 \to -5x=-60 \to x=12$
$y=30-2(12)=6$: $P=10(12)+7(6)=120+42=162$

Step5: Compare profit values

$162 > 150 > 105 > 0$

Answer:

D. Ms. Waller should make 12 posy bouquets and 6 cascade bouquets to maximize her profit.