Chapter 7.6, Problem 43E

To determine

The optimal acreage for each fruit.

Answer to Problem 43E

The maximum acreage is 60 acres for crop A, 90 acres for crop B.

Explanation of Solution

Given:

The Yield per acre on crop A is 300; yield per acre for crop B is 500

Formula used:

Solving the equations based on substitution method.

Calculation:

Let $x$ acres of land increased for crop A and $y$ acres of land for crop B

Yield per acre on crop A is 300; yield per acre for crop B is 500

Then the objective function is

  $Y=300x+500y$

Total 150 acres of land for raising the crop, that is

  $x+y\le 150$

Number of days takes to trim crop A is $x$

Number of days takes to trim crop B is $2y$

Total available days is 240, that is

  $x+2y\le 240$

Number of days takes to pick crop A is $0.3x$

Number of days takes to pick crop B is $0.1y$

Total available days is 30, that is

  $0.3x+0.1y\le 30$

Now the given data as a linear programming problem is

Objective function: $Y=300x+500y$

The constraints are given below as,

  $\begin{array}{l}x\ge 0\\ y\ge 0\\ x+y\le 150\\ x+2y\le 240\\ 0.3x+0.1y\le 30\end{array}$

First draw the region corresponding to the system of constraints

  

Here vertices of the region are $\left(0,0\right),\left(100,0\right),\left(75,75\right),\left(60,90\right),\left(0,120\right)$

The objective function at each vertex is given below as,

  $\begin{array}{l}Y\left(0,0\right)=300\left(0\right)+500\left(0\right)\\ =0\\ Y\left(100,0\right)=300\left(100\right)+500\left(0\right)\\ =30,000\\ Y\left(0,120\right)=300\left(0\right)+500\left(120\right)\\ =63,000\\ Y\left(75,75\right)=300\left(75\right)+500\left(75\right)\\ =60,000\end{array}$

Hence the maximum yield is $63,000$

Maximum acreage is $\begin{array}{l}60\text{acres}\text{for}\text{crop}A\\ 90\text{acres}\text{for}\text{crop}B\end{array}$

Conclusion:

The maximum acreage is 60 acres for crop A, 90 acres for crop B.