Answer to Question #136818 in Operations Research for Cameron Foster

Question #136818
Mr. Trenga plans to start a new business called River Explorers, which will rent canoes and kayaks
to people to travel 10 miles down the Clarion River in Cook Forest State Park. He has $45,000 to
purchase new boats. He can buy the canoes for $600 each and the kayaks for $750 each. His facility
can hold up to 65 boats. The canoes will rent for $25 a day, and the kayaks will rent for $30 a day. How
many canoes and how many kayaks should he buy to earn the most revenue? What is the maximum
revenue
1
Expert's answer
2020-10-06T16:24:50-0400

 Let, number of cаnoes bought = "x"

Number of kayaks bought = "y"

The optimization function is:

"Z = 25x + 30y"

Constraints are:

"600*x + 750*y \\leq 45000"

"x + y \\leq 65"

Since the number of boat can`t be negative,

"x \\geq 0,\\\\ y \\geq 0"


The linear programming problem is:

 "Z = 25*x + 30*y"

subject to:

"600*x + 750*y \\leq 45000"

"x + y \\leq 65"

"y \\geq 0"

"x \\geq 0;"


Graph:



First equation:

"600*x + 750*y = 45000"

Divide both parts by 150

"4*x + 5*y = 300"


Second equation:

"x + y = 65"


Let's solve the system of equations:

"\\begin{cases}\n 4*x + 5*y = 300 \\\\\nx + y \\leq 65\n\\end{cases}"


From the first equation we express x:

"4*x+5*y= 300\\\\\n4*x = 300-5*y\\\\\n(4*x)\/4 = (300-5*y)\/4\\\\\nx=75-(5*y)\/4"


Substitute the found x into the second equation "x + y = 65" ;


We get:

"y + (75 - (5*y)\/4) =65\\\\\n75-y\/4 = 65\\\\\n-y\/4=65-75\\\\\ny\/4 - 10\\\\\ny = 40"


Substitute the found y into the equation:

"x=75-(5*40)\/4\\\\\nx = 75-50\\\\\nx = 25"


On solving these we get:

"x =" number of canoes "= 25"

"y =" number of kayaks "= 40"


Z "= ( 25 \u00d7 25 ) + ( 30 \u00d7 40 ) = 1825"


Answer:

Mr. Trenga should buy 25 canoes and 40 kayaks to earn the most revenue.

Maximum revenue is $1825.


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