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A farmer wants to maximize profit from selling onions and greens. Onions yield 500,000 profit/ton and greens yield 300,000 profit/ton. The farmer has 10 hectares of land (onions: 1 ha/ton, greens: 2 h

Mathematics

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Step-by-step solution

4 steps

Here's the solution to your linear programming problem:

Task (a):

  1. Define variables.

    Let xx be the number of tons of onions.
    Let yy be the number of tons of greens.

  2. Formulate the objective function.

    The farmer wants to maximize profit.
    Profit from onions: Ugx. 500,000 per ton.
    Profit from greens: Ugx. 300,000 per ton.
    The objective function to maximize is:

    P=500,000x+300,000yP = 500,000x + 300,000y
  3. Formulate the constraints.

    • Land Constraint: The farmer has "almost ten hectares" of land. Onions need 1 hectare per ton, and greens need 2 hectares per ton. 1x+2y≤101x + 2y \le 10
    • Labor Constraint: The farmer intends to spend "not more than 120 hours per week" on labor. Onions require 2 hours per ton, and greens require 3 hours per ton. 2x+3y≤1202x + 3y \le 120
    • Non-negativity Constraints: The number of tons of crops cannot be negative. x≥0x \ge 0 y≥0y \ge 0

    The required inequalities are:

    x+2y≤10\boxed{x + 2y \le 10} 2x+3y≤120\boxed{2x + 3y \le 120} x≥0\boxed{x \ge 0} y≥0\boxed{y \ge 0}

    Task (b):

  4. Graph the inequalities and identify the feasible region.

    To graph the inequalities, we consider their boundary lines:

    1. For x+2y=10x + 2y = 10: If x=0x=0, 2y=10⇒y=52y=10 \Rightarrow y=5. Point: (0,5)(0, 5). If $y=
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