Question

because of limited materials a skate manufacturer can produce no more than 60 pairs of hockey skates, and 40 pairs of figure skates, per day. the company can make 70 or more skates in total per day. it costs $120 to make a hockey skate and $100 to make a figure skate. the company wants to know what combination will result in the minimum and maximum costs and what that cost will be.

Explanation:

Step1: Define variables

Let $x$ be the number of hockey - skates and $y$ be the number of figure - skates.

Step2: Set up constraints

We have the following constraints:

1. $x\leq60$ (limit on hockey - skates production)
2. $y\leq40$ (limit on figure - skates production)
3. $x + y\leq70$ (total production limit)
4. $x\geq0,y\geq0$ (non - negativity constraints)

Step3: Define the cost function

The cost function $C = 120x+100y$

Step4: Find the corner points of the feasible region

• Intersection of $x = 0$ and $y = 0$: Point $(0,0)$
• Intersection of $x = 0$ and $y=40$: Point $(0,40)$
• Intersection of $x = 60$ and $y = 0$: Point $(60,0)$
• Intersection of $x + y=70$ and $x = 60$: Substitute $x = 60$ into $x + y=70$, we get $y = 10$, so the point is $(60,10)$
• Intersection of $x + y=70$ and $y = 40$: Substitute $y = 40$ into $x + y=70$, we get $x = 30$, so the point is $(30,40)$

Step5: Evaluate the cost function at the corner points

• At $(0,0)$: $C(0,0)=120\times0 + 100\times0=0$
• At $(0,40)$: $C(0,40)=120\times0+100\times40 = 4000$
• At $(60,0)$: $C(60,0)=120\times60+100\times0 = 7200$
• At $(60,10)$: $C(60,10)=120\times60 + 100\times10=7200 + 1000=8200$
• At $(30,40)$: $C(30,40)=120\times30+100\times40=3600 + 4000=7600$

Answer:

The minimum cost is $0$ (when no skates are produced) and the maximum cost is $8200$ (when 60 hockey - skates and 10 figure - skates are produced)