a company produces three types of shoes, formal, casual, and athletic, at its two factories; factory i and…

a company produces three types of shoes, formal, casual, and athletic, at its two factories; factory i and factory ii. the company must produce at least 6000 pairs of formal shoes, 8000 pairs of casual shoes, and 9000 pairs of athletic shoes. daily production of each factory for each type of shoe is:\n| |factory i|factory ii|\n|--|--|--|\n|formal|100|100|\n|casual|100|200|\n|athletic|300|100|\noperating factory i costs $1500 per day and it costs $2000 per day to operate factory ii. what is the minimum cost to operate both factories? round your answer to the nearest whole number. do not include a dollar sign or comma in your answer.

a company produces three types of shoes, formal, casual, and athletic, at its two factories; factory i and factory ii. the company must produce at least 6000 pairs of formal shoes, 8000 pairs of casual shoes, and 9000 pairs of athletic shoes. daily production of each factory for each type of shoe is:\n| |factory i|factory ii|\n|--|--|--|\n|formal|100|100|\n|casual|100|200|\n|athletic|300|100|\noperating factory i costs $1500 per day and it costs $2000 per day to operate factory ii. what is the minimum cost to operate both factories? round your answer to the nearest whole number. do not include a dollar sign or comma in your answer.

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of days Factory I operates and $y$ be the number of days Factory II operates.

Step2: Set up constraints

For formal shoes: $100x + 100y\geq6000$, simplifies to $x + y\geq60$. For casual shoes: $100x+200y\geq8000$, simplifies to $x + 2y\geq80$. For athletic shoes: $300x+100y\geq9000$, simplifies to $3x + y\geq90$. Also, $x\geq0,y\geq0$.

Step3: Set up cost function

The cost function $C=1500x + 2000y$.

Step4: Solve the system of inequalities graphically or using linear - programming methods

First, find the intersection points of the boundary lines of the inequalities: Intersection of $x + y=60$ and $x + 2y=80$: Subtract the first equation from the second: $(x + 2y)-(x + y)=80 - 60$, so $y = 20$ and $x=40$. Intersection of $x + y=60$ and $3x + y=90$: Subtract the first equation from the second: $(3x + y)-(x + y)=90 - 60$, $2x=30$, $x = 15$ and $y=45$. Intersection of $x + 2y=80$ and $3x + y=90$: From $x + 2y=80$, we have $x=80 - 2y$. Substitute into $3x + y=90$: $3(80 - 2y)+y=90$, $240-6y + y=90$, $- 5y=90 - 240=-150$, $y = 30$ and $x=20$.

Step5: Evaluate the cost function at intersection points

At $(40,20)$: $C=1500\times40+2000\times20=60000 + 40000=100000$. At $(15,45)$: $C=1500\times15+2000\times45=22500+90000=112500$. At $(20,30)$: $C=1500\times20+2000\times30=30000+60000=90000$.

Answer:

90000