Answer :
Maximize linear programming problem $6R + $8C subject to 4R + 5C ≤ 10,000, 5R + 2C ≤ 5,000, R ≥ 500, R ≥ 0, and C ≥ 0.
Formulate a linear programming problem to maximize profit by determining the optimal quantities of roses (R) and carnations (C) to be grown, given the constraints 4R + 5C ≤ 10,000, 5R + 2C ≤ 5,000, R ≥ 500, R ≥ 0, and C ≥ 0?
Decision variables:
Let's define the decision variables for this problem:
Let R represent the number of bunches of roses to be grown.
Let C represent the number of bunches of carnations to be grown.
Linear program formulation:
Objective function:
Maximize profit: $6R + $8C
Constraints:
Area constraint: The total area used for planting cannot exceed the available 10,000 square feet.
4R + 5C ≤ 10,000
Fertilizer constraint: The total amount of fertilizer used cannot exceed the available 5,000 pounds.
5R + 2C ≤ 5,000
Sales commitment constraint: At least 500 bunches of roses need to be grown.
R ≥ 500
Non-negativity constraint: The number of bunches of roses and carnations cannot be negative.
R ≥ 0, C ≥ 0
Mathematically, the linear program can be written as:
Maximize 6R + 8C
Subject to:
4R + 5C ≤ 10,000
5R + 2C ≤ 5,000
R ≥ 500
R ≥ 0, C ≥ 0
To maximize profit, the objective function is defined as the sum of the profit contributions from roses and carnations, which is $6R + $8C. The goal is to determine the optimal values of R and C that maximize this objective.
The area constraint ensures that the total area used for planting (4 square feet per bunch of roses and 5 square feet per bunch of carnations) does not exceed the available 10,000 square feet.
The fertilizer constraint ensures that the total amount of fertilizer used (5 pounds per bunch of roses and 2 pounds per bunch of carnations) does not exceed the available 5,000 pounds.
The sales commitment constraint requires that at least 500 bunches of roses be grown, ensuring the fulfillment of sales commitments.
The non-negativity constraints state that the number of bunches of roses and carnations cannot be negative since we cannot have negative quantities.
By solving this linear program, the optimal values of R and C can be obtained, representing the number of bunches of roses and carnations to be grown, respectively, in order to maximize profit.
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