Question

1. problem: a snack bar cooks and sells hamburgers and hot dogs during football games. the snack bar cannot cook more than 90 items total. they must sell at least 20 hamburgers and at least 10 hot dogs because the meat is going to expire in a few days. the profit on a hamburger is $0.33 and the profit on a hot dog is $0.21. how many of each item should it sell to make the maximum profit?

a) objective function: (2 pts)
b) constraints: (3 pts)
c) vertices of feasible region: (3 pts)
d) how many hot dogs and how many hamburgers need to be sold to make the maximum profit? (4 pts)

Explanation:

Step1: Define Variables

Let \( x \) = number of hamburgers, \( y \) = number of hot dogs.

Step2: Objective Function (a)

Profit from hamburgers: \( 0.33x \), from hot dogs: \( 0.21y \).
Objective: Maximize \( P = 0.33x + 0.21y \).

Step3: Constraints (b)

1. Total items: \( x + y \leq 90 \)
2. Hamburgers: \( x \geq 20 \)
3. Hot dogs: \( y \geq 10 \)
4. Non - negativity: \( x \geq 0, y \geq 0 \) (already covered by \( x\geq20,y\geq10 \))

Step4: Feasible Region Vertices (c)

• Intersection of \( x = 20 \), \( y = 10 \): \( (20,10) \)
• Intersection of \( x = 20 \), \( x + y = 90 \): \( y=90 - 20=70\), so \( (20,70) \)
• Intersection of \( y = 10 \), \( x + y = 90 \): \( x=90 - 10 = 80\), so \( (80,10) \)

Step5: Evaluate Profit at Vertices (d)

• At \( (20,10) \): \( P=0.33(20)+0.21(10)=6.6 + 2.1 = 8.7 \)
• At \( (20,70) \): \( P=0.33(20)+0.21(70)=6.6+14.7 = 21.3 \)
• At \( (80,10) \): \( P=0.33(80)+0.21(10)=26.4 + 2.1 = 28.5 \)

Answer:

a) Objective Function: \( \boldsymbol{P = 0.33x + 0.21y} \) (to maximize)
b) Constraints: \( \boldsymbol{x + y \leq 90} \), \( \boldsymbol{x \geq 20} \), \( \boldsymbol{y \geq 10} \)
c) Vertices: \( \boldsymbol{(20,10)} \), \( \boldsymbol{(20,70)} \), \( \boldsymbol{(80,10)} \)
d) To maximize profit, sell \( \boldsymbol{80} \) hamburgers and \( \boldsymbol{10} \) hot dogs.