Answer to Question #281298 in Operations Research for Benson

Question

A farmer can plant upto to 8 acres of land with wheat and barley. He can earn $5000 for every acre he plants with wheat and $3000. His use of necessary pesticides is limited by federal regulations of 10 gallons for his entire 8 acres. Wheat requires 2 gallons of acre for each planted and barley one gallon per acre. What is the maximum profit he can make

Answer 1

Solution:

let x= the number of acres of wheat

let y= the number of acres of barley.

since the farmer earns 5,000 for each acre of wheat and 3,000 for each acre of barley, then the total prof it the farmer can earn is $5000\times x+3000\times y$ .

let p= total profit that can be earned, your equation for profit becomes:

$p=5000 x+3000 y$

that's your objective function. it's what you want to maximize

the constraints are:

number of acres has to be greater than or equal to 0 .

number of acres has to be less than or equal to 8 .

amount of pesticide has to be less than or equal to 10 .

your constraint equations are:

$\begin{aligned} &x\ge0 \\ &y\ge0 \\ &x+y\le8 \\ &2 x+y\le10 \end{aligned}$

to graph these equations, solve for y in those equations that have y in them and then graph the equality portion of those equations.

$\begin{aligned} &x\ge0 \\ &y\ge0 \\ &y\le8-x \\ &y\le10-2 x \end{aligned}$

x=0 is a vertical line that is the same line as the y-axis.

y=0 is a horizontal line that is the same line as the x-axis.

the area of the graph that satisfies all the constraints is the region of feasibility.

the maximum or minimum solutions to the problem will be at the intersection points of the lines that bound the region of feasibility. The graph of your equations looks like this:

Thus, maximum profit is $28,000 when x=2, y=6.