Q40. - Algebra 2

Question

Dean Stadler has 20 days in which to plant corn and soybeans. The corn can be planted at a rate of 250 acres per day and the soybeans at a rate of 200 acres per day. He has 4500 acres available for planting these two crops.

 

40. If the profit on corn is \(26 per acre and the profit on soybeans is \)30 per acre, how much of each should Mr. Stadler plant? What is the maximum profit?

Step-by-Step Solution

Answer

Hence, he should plant 0 acres of corn and 20 acres of soybeans to get the maximum profit of $600.

Step 1 – Construct a system of equations for the given situation.

Let c and s be the number of acres of corn and soybean respectively.

Given that Dean Stadler wants to plant corn and soybeans in 20 days.

So, $c+s\le 20$.

The corn can be planted at a rate of 250 acres per day and the soybeans at a rate of 200 acres per day. 

He has 4500 acres available for planting these two crops.

So, $250c+200s\le 4500\Rightarrow 5c+4s\le 90$.

Also, $c\ge 0,s\ge 0.$

Then, the system of equations is:

  $\begin{array}{c}c+s\le 20\\ 5c+4s\le 90\\ c\ge 0\\ s\ge 0\end{array}$.

Step 2 – Find the intersecting point of c + s = 20 , 5 c + 4 s = 90

To find the intersecting point, use the elimination method.

Multiply the equation $c+s=20$ by 4 and subtract the new resultant equation from $5c+4s=90$

 $\begin{array}{l}\underset{¯}{\begin{array}{c}5c+4s=90\\ c+s=20\end{array}}\begin{array}{c}\\ \text{multiply by 4}\end{array}\underset{¯}{\begin{array}{c}5c+4s=90\\ \left(-\right)4c+4s=80\end{array}}\\ c+0=10\end{array}$

So, $c=10$

Substitute $c=10$ in $c+s=20$ and solve for s :

 $\begin{array}{c}c+s=20\\ 10+s=20\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}substitute 10 for }c\\ s=10\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}subtract 10 from both sides}\end{array}$

So, the intersecting point is $\left(10,10\right).$

Step 3 – Graphical representation of the above system of inequality

Graph the solution region for $c+s\le 20$ and the solution region for $5c+4s\le 90$.

The intersecting region of $c+s\le 20$and $5c+4s\le 90$is the solution region of the system of inequalities.

The feasible region of the system of equations is shaded in purple color, and the coordinates of the vertices of the feasible region are $\left(0,0\right),\left(0,20\right),\left(18,0\right),\left(10,10\right).$

Step 4 – Construct the profit function and evaluate at the coordinates of the vertices of the feasible region.

Given that the profit on corn is $26 per acre and the profit on soybeans is $30 per acre.

So, profit $P=26c+30s$.

 

Find $P=26c+30s$ at $\left(0,0\right),\left(0,20\right),\left(18,0\right),\left(10,10\right).$

 

Substitute $\left(c,s\right)=\left(0,0\right)$ in $P=26c+30s$:

 $\begin{array}{c}P=26\left(0\right)+30\left(0\right)\\ P=0\end{array}$

Substitute $\left(c,s\right)=\left(0,20\right)$in $P=26c+30s$:

 $\begin{array}{c}P=26\left(0\right)+30\left(20\right)\\ P=600\end{array}$

 

Substitute $\left(c,s\right)=\left(18,0\right)$in $P=26c+30s$:

 $\begin{array}{c}P=26\left(18\right)+30\left(0\right)\\ P=468\end{array}$

 

Substitute $\left(c,s\right)=\left(10,10\right)$in $P=26c+30s$:

 $\begin{array}{c}P=26\left(10\right)+30\left(10\right)\\ P=560\end{array}$

 

From the calculated values, one can conclude that the maximum of P is 600.

Hence, he should plant 0 acres of corn and 20 acres of soybeans to get the maximum profit of $600.