A man owns a field of area 1000 sq.m. He wants to plant fruit trees in it. He has a sum of Rs 1400 to purchase young trees. He has the choice of two types of trees. Type *A *requires 10 sq.m of ground per tree and costs Rs 20 per tree and type *B* requires 20 sq.m of ground per tree and costs Rs 25 per tree. When fully grown, type *A* produces an average of 20 kg of fruit which can be sold at a profit of Rs 2.00 per kg and type *B *produces an average of 40 kg of fruit which can be sold at a profit of Rs. 1.50 per kg. How many of each type should be planted to achieve maximum profit when the trees are fully grown? What is the maximum profit?

#### Solution

Let the man planted *x* trees of type *A* and *y* trees of type *B*.

Number of trees cannot be negative.

Therefore, \[x, y \geq 0\] To plant tree of type *A* requires 10 sq.m and type *B* requires 20 sq.m of ground per tree. And, it is given that a man owns a field of area 1000 sq.m.Therefore,

\[10x + 20y \leq 1000\]

Type *A* costs Rs 20 per tree and type *B* costs Rs 25 per tree. Therefore, *x* trees of type *A*and *y* trees of type *B *costs Rs 20*x** *and Rs 25*y** *respectively. A man has a sum of Rs 1400 to purchase young trees.

\[20x + 25y \leq 1400\]

Thus, the mathematical formulation of the given linear programmimg problem is

Max Z = 40*x *− 20*x *+ 60*y *− 25*y* = 20*x* + 35*y*

subject to

\[10x + 20y \leq 1000\]

\[20x + 25y \leq 1400\]

The feasible region determined by the system of constraints is

The corner points are A(0, 50), B(20, 40), C(70, 0)

The values of Z at these corner points are as follows

Corner point | Z = 20x + 35y |

A | 1750 |

B | 1800 |

C | 1400 |

The maximum value of Z is 1800 which is attained at B(20, 40)

Thus, the maximum profit is Rs 1800 obtained when Rs 20 were invested on type A and Rs 40 were invested on type B.