Question

A farmer is going to divide her 60 acre farm between two crops. Seed for crop A costs $20 per acre. Seed for crop B costs $40 per acre. The farmer can spend at most $2200 on seed.

If crop B brings in a profit of $320 per acre, and crop A brings in a profit of $140 per acre, how many acres of each crop should the farmer plant to maximize her profit?

Answer 1

Let X = acres for A  and  Y = acres for B.


The farmer wants to maximize his profit  

    P(X,Y) = 140*X + 320*Y


under these restrictions

    X + Y = 60,

    20*X + 40*Y <= 2200,

    X >= 0,  Y >= 0.



The feasible domain is shown in the Figure below.

Plot X + Y = 60 (red)  and  20X + 40Y = 2200 (green)



The feasibility domain is the quadrilateral in QI below the red and green lines.


Corner points of the quadrilateral (its vertices) are

    P1 = (0,0)

    P2 = (0,55)   (y-intercept of the green line)

    P3 = (10,50)  (intersection point of red and green line).

    P4 = (60,0)    (x-intercept of the red line)


The values of the profit function in the corner points are:


    P1:  P( 0, 0) = 0

    P2:  P( 0,55) = 140*0  + 320*55 = 17600

    P3:  P(10,50) = 140*10 + 320*50 = 17400

    P4:  P(60,0)  = 140*60 + 320*0  = 8400.


The maximal value of the profit function is achieved at the points P2.

It means that the optimal solution is 0 acres for A and 55 acres for B.

Hope This Helps !