Question

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

Answer 1

 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:

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

From the first equation we express x:

$4*x+5*y= 300\\ 4*x = 300-5*y\\ (4*x)/4 = (300-5*y)/4\\ x=75-(5*y)/4$

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

We get:

$y + (75 - (5*y)/4) =65\\ 75-y/4 = 65\\ -y/4=65-75\\ y/4 - 10\\ y = 40$

Substitute the found y into the equation:

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

On solving these we get:

$x =$ number of canoes $= 25$

$y =$ number of kayaks $= 40$

Z $= ( 25 × 25 ) + ( 30 × 40 ) = 1825$

Answer:

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

Maximum revenue is $1825.