1. A farmer has 10 acres to plant in wheat and rye. He has to plant at least 7 acres. However, he has only \$1200 to spend and each acre of wheat costs \$200 to plant and each acre of rye costs \$100 to plant. Moreover, the farmer has to get the planting done in 12 hours and it takes an hour to plant an acre of wheat and 2 hours to plant an acre of rye. If the profit is \$500 per acre of wheat and \$300 per acre of rye how many acres of each should be planted to maximize profits?

Steps

0. Read the whole problem.
1. Define your unknowns.
2. Express the objective function and the constraints.
3. Graph the constraints.
4. Find the corner points to the region of feasible solutions.
5. Evaluate the objective function at all the feasible corner points.

### 1. Define your unknowns

In this case the question is how many acres of wheat and how many acres of rye should be planted.

Let   x = the number of acres of wheat

and   y = the number of acres of rye.

### 2. Express the objective and the constraints.

First, the objective. The object is to maximize profits. Each acre of wheat brings in \$500 and each acre of rye brings in \$300.

Profit = 500x + 300y

After getting the unknowns defined and the objective expressed, everything else is a constraint.

The first constraint is the acreage constraint. The farmer has only 10 acres to plant. So the total number of acres has to be less than or equal to 10. There is another acreage constraint. The farmer has to plant at least 7 acres.

x + y ≤ 10

x + y ≥ 7

The next constraint is the cost constraint. The farmer has only \$1200 to spend on the planting.

200x + 100y ≤ 1200

Finally there is the time constraint. The planting has to be done in 12 hours.

x + 2y ≤ 12

and of course the implied constraints

x ≥ 0,   and  y ≥ 0

### 3. Graph the constraints.

The implied constraints tell us that the feasible solutions will all lie in the upper right quadrant of the plane.

First we find the intercepts

 Let us first graph the acreage constraints. The first acreage constraint has 10 for both the x and y intercepts.

 Plot 10 on both the x and y axes, and connect the points with a straight line. Next decide which side of the line is the side that contains solutions to the inequality. Pick a point which is not on the line. In this case the origin is the simplest such point with which to work. Does the origin satisfy this constraint?

At the origin, no acres are planted. Does the farmer have enough acres for this? The answer is yes. So the origin is on the right side of this line. Shade in all of the points in the upper right quadrant which are on the same side of the line as the origin.

The lower acreage constraint has x and y intercepts which are both equal to 7. In this case, the origin is not a solution to this constraint. At the origin, the farmer is not planting any acres of anything, but this constraint says that the farmer needs to plant at least 7 acres. If the origin is not on the correct side, the other side is the correct side. The region of feasible solutions has been cut down to the shaded area above.

In general, if both unknowns are on the same side and all the coefficients are positive, the line will have positive x and y intercepts, and the side toward the origin will be the less than side and the side away from the origin will be the greater than side.

Next graph the cost constraint. It has an x intercept of 6 and a y intercept of 12.

 Since this is a "less than" constraint the side on the same side as the origin is shaded.

Finally there is the time constraint. It has an x intercept of 12 and a y intercept of 6.

 ### 4. Find the feasible corner points

The feasible corner points are the ones at the corners of the shaded region. To find them we identify the equations which are associated with the lines that intersect at the points.

 We solve these systems involving the equations of the lines, which meet at these corner points, simultaneously.

x = 5

We substitute this into x + y = 7 and get

y = 2

so

A = (5, 2)

 x = 4

We substitute this into x + 2y = 12

(4) + 2y = 12

Subtract 4 from both sides.

2y = 8

Divide by 2

y = 4

so

B = (4, 4)

 If we subtract the two equations we get

y = 5

If we substitute that into   x + y = 7,   we get

x = 2

so

C = (2, 5)

We see that none of the feasible corner points is on the upper acreage constraint, the fact that the farmer had only 10 acres. It turns out to be what is called a superfluous constraint. The most number of acres that can be planted is 8 by planting 4 acres of each crop. If all of the other constraints are satisfied, the farmer will automatically be planting less than 10 acres.

### 5. Evaluate the objective function in all of the feasible corner points.

A = (5, 2)

Profit = \$500(5) + \$300(2) = \$2500 + \$600 = \$3100

B = (4, 4)

Profit = \$500(4) + \$300(4) = \$2000 + 1200 = \$3200

C = (2, 5)

Profit = \$500(2) + \$300(5) = \$1000 + \$1500 = \$2500

The winner is B:

The maximize the profit of \$3200 is obtained by planting 4 acres of each crop.

If we think about it, this makes sense.   C   can be seen to be the least profitable corner point. You are planting the fewest number of acres and you are planting more of the less profitable rye and less of the more profitable wheat. The real question is whether  A   or   B   will be more profitable. At   B   you are planting more acres than at   A,   but at   A,   you are planting more of the more profitable wheat and less of the less profitable rye. One way to decide is to actually compute the profit at both points and see which one would be best.

top

Math 131

Steve Wilson