a small organic food store makes two types of fruit smoothies: tropical and sport. each tropical smoothie…

a small organic food store makes two types of fruit smoothies: tropical and sport. each tropical smoothie uses 9 oz of orange juice and 6 oz of milk. each sport smoothie uses 3 oz of orange juice and 3 oz of milk. the store buys the orange juice in a 210-oz container and milk in a 150-oz container. the store sells the tropical smoothies for $4 and sells the sport smoothie for $3. what is the maximum in smoothie sales the store can make from 1 container each of orange juice and milk?\n$100\n$110\n$150\n$280

a small organic food store makes two types of fruit smoothies: tropical and sport. each tropical smoothie uses 9 oz of orange juice and 6 oz of milk. each sport smoothie uses 3 oz of orange juice and 3 oz of milk. the store buys the orange juice in a 210-oz container and milk in a 150-oz container. the store sells the tropical smoothies for $4 and sells the sport smoothie for $3. what is the maximum in smoothie sales the store can make from 1 container each of orange juice and milk?\n$100\n$110\n$150\n$280

Answer

Explanation:

Step1: Define variables

Let $x$ = number of tropical smoothies, $y$ = number of sport smoothies.

Step2: Set up constraints

Orange juice constraint:

Each tropical uses 9oz, sport uses 3oz; total 210oz. $9x + 3y \leq 210$ Simplify: $3x + y \leq 70$

Milk constraint:

Each tropical uses 6oz, sport uses 3oz; total 150oz. $6x + 3y \leq 150$ Simplify: $2x + y \leq 50$

Non-negativity constraints:

$x \geq 0$, $y \geq 0$

Step3: Set up objective function

Profit $P = 4x + 3y$ (maximize this)

Step4: Find feasible region vertices

  1. Intersection of $3x + y = 70$ and $2x + y = 50$: Subtract the second equation from the first: $(3x + y) - (2x + y) = 70 - 50$ $x = 20$ Substitute $x=20$ into $2x + y = 50$: $2(20) + y = 50$ $y = 10$ Vertex 1: $(20, 10)$
  2. Intersection of $3x + y = 70$ and $x=0$: $y = 70$, Vertex 2: $(0, 70)$
  3. Intersection of $2x + y = 50$ and $y=0$: $2x = 50$ → $x=25$, Vertex 3: $(25, 0)$
  4. Origin: $(0, 0)$

Step5: Evaluate profit at vertices

  1. At $(20, 10)$: $P = 4(20) + 3(10) = 80 + 30 = 110$
  2. At $(0, 70)$: $P = 4(0) + 3(70) = 210$ (invalid, violates milk constraint $2(0)+70=70>50$)
  3. At $(25, 0)$: $P = 4(25) + 3(0) = 100$
  4. At $(0, 0)$: $P = 0$

Answer:

$110