Question

Select the correct answer from each drop-down menu.

A farmer plans to grow apples on part of his land and peaches on another. He can afford a maximum of 54 acres of land for the two. In the summer, the apple orchard will require 3,000 gallons of water per acre each day, and the peach orchard will require 800 gallons of water per acre each day. However, his irrigation system can deliver a maximum of 80,000 gallons a day.
From the apple orchard, he expects to earn a profit of $3,400 per acre each year. From the peach orchard, he expects a profit of $1,600 per acre each year. What division of land will maximize his expected profit?
One of the graphs below shows the system of inequalities that represents this situation with its solution region shown as the overlap of the shaded regions.

Answer the following questions.
Which graph shows the inequality that represents this situation with the solution region shown as the overlap of the shaded regions? ____ (Graph A, Graph B, Graph C, Graph D)
Will the farmer be able to maximize his profits if the apple orchard takes up 30 acres and the peach orchard takes up 20 acres?
____ (Yes, No), because the point (30,20) lies ____ (Inside, Outside) the solution region.

Answer 1

Graph B, No, Outside

Explanation:

Which graph shows the inequality that represents this situation with the solution region shown as the overlap of the shaded regions?

Graph B

Will the farmer be able to maximize his profits if the apple orchard takes up 30 acres and the peach orchard takes up 20 acres?

No , because the point (30,20) lies outside the solution region.

Answer 2

Graph B

No

Inside

Explanation:

Let, the number of apples = x and number of peaches = y.

It is given that the farmer can afford maximum of 54 acres.

Thus, x + y ≤ 54.

Also, apples require 3000 gallons of water and peaches require 800 gallons of water each day.

Since, maximum amount of water is 80,000 gallons per day,

We get, 3000x + 800y ≤ 80,000.

It is required to maximize the profit given by z = 3400x + 1600y.

Thus, the system of equations becomes,

z = 3400x + 1600y

3000x + 800y ≤ 80,000.

x + y ≤ 54.

Plotting these equations gives the following graph.

We can see that the graph obtained is equivalent to Graph B.

Further, we know that the maximum value of the objective function z = 3400x + 800y is obtained at the boundary point of the solution region.

Thus, any point inside the solution region will give a feasible solution not maximum solution.

Hence, ( 30,20 ) will not maximize the profit as it lies inside the solution region.