Q. 20 - Precalculus Enhanced with Graphing Utilities

Question

Farm Management A farmer has 70 acres of land available for planting either soybeans or wheat. The cost of preparing the soil, the workdays required, and the expected profit per acre planted for each type of crop are given in the following table:

The farmer cannot spend more than $\backslash (1800$ in preparation costs nor use more than a total of 120 workdays. How many acres of each crop should be planted to maximize the profit? What is the maximum profit? What is the maximum profit if the farmer is willing to spend no more than $\backslash )2400$ on preparation? 

Step-by-Step Solution

Answer

The maximum profit will be obtained if the number of acres of soya beans is 40 and number of acres of wheat is 0 . The maximum profit obtained from the crops is $\$7200$. 

Step 1. Given information

Let the number acres of soya beans and wheat on the land be x and y respectively.
The farmer cannot spend more than $\$1800$ on preparation costs. The preparation cost for soya beans is $\$60$ per acre and preparation cost for wheat is $\$30$ per acre. These conditions can be written in terms of x and y.

$60x+30y\le 1800$ 

Step 2.The number of acres cannot be less than zero;

Therefore the following inequalities are also required to be satisfied.

$x\ge 0$

The farmer is not ready to use more than 120 workdays for the work. The workdays required for soya beans is 3 days per acre and the workdays required for wheat is 4 days per acre. These conditions can be written in terms of x and y .

$3x+4y\le 120$

The total acres of land available to the farmer for plantation of wheat and soya beans is no more than 70 acres.

$x+y\le 70$

The profit obtained per acre from planting soya beans and wheat is $\$180$ per acre and $\$100$ per acre respectively.

$P=180x+100y$

The region where we can evaluate the maximum of the profit function $P=180x+100y$ can be obtained by graphing the inequalities.

Step 3. Finding the value of P

We need to determine the value of the expression $P=180x+100y$ at each vertex in order to determine the maximum of the expression. We have given below the table of the values calculated at each of the vertex.

Vertex	Value of $P=180x+100y$
$(30,0)$	$P=180\left(30\right)+100\left(0\right)=5400$
$(0,30)$	$P=180\left(0\right)+100\left(30\right)=3000$
$(24,12)$	$P=180\left(24\right)+100\left(12\right)=5520$

Therefore, the maximum profit will be obtained if the number of acres of soya beans is 24 and number of acres of wheat is 12 . The maximum profit obtained from the crops is $\$5520$.
The farmer is now ready to pay $\$2400$ in total for the preparation of soil; the inequalities obtained for this case will be given as follows:

$$\begin{gathered} 60x+30y\le 2400 \\ 3x+4y\le 120 \\ x+y\le 70 \\ x\ge 0 \\ y\ge 0 \end{gathered}$$

Step 4. The region where we can evaluate the maximum of the profit function P = 180 x + 100 y can be obtained by graphing the inequalities.

We need to determine the value of the expression $P=180x+100y$ at each vertex in order to determine the maximum of the expression. We have given below the table of the values calculated at each of the vertex.

Vertex	Value of $P=180x+100y$
$(40,0)$	$P=180\left(40\right)+100\left(0\right)=7200$
$(0,30)$	$P=180\left(0\right)+100\left(30\right)=3000$

Therefore, the maximum profit will be obtained if the number of acres of soya beans is 40 and number of acres of wheat is 0 . The maximum profit obtained from the crops is $\$7200$.