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- Solving a linear programming related to quantities of plants

Hello,

In grade 11 of high school, I encountered this linear programming problem on my textbook:

Every six months, a decorative plants seller order Aglanonemas (A) and Sansevierias (S) to an agent, which respectively give profit of $500 and $350. The agent requires that plant A must be ordered by minimum of 20% from total plant orders. However, the seller have garden which is only sufficient for 10 As and 15 Ss. Assumed that all plants are sold out and the profit is maximized. How many As and Ss should be ordered?

The "alternative solution" described in the textbook as follows:

Let:

- ##x## : amount of plant A

- ##y## : amount of plant S

- ##L## : garden area

- ##L_x## : area of garden for one plant A

- ##L_y## : area of garden for one plant B

Since the plant order requirement is 20% of total orders are for plant A, we write: $$x \geq \frac 1 5 \left( x + y \right) \text{or } 4x - y \geq 0$$

However, the garden capacity is only for 10 A plants and 15 S plants, that is $$ x \cdot (\frac 1 {10} L) + y \cdot (\frac 1 {15} L \leq L) \text{ or } 3x + 2y \leq 30$$

The objective function is ##Z = 5x + 3.5y## (in the order of $100).

The complete linear problem equations are:

\begin{cases}

4x - y \geq 0 \\

3x + 2y \leq 30 \\

x \geq 0 \\

y \geq 0 \\

Z = 5x + 3.5y \text{ } \text{(maximize)}

\end{cases}

Solving the problem (using graphical method), the solutions are: ##B(2\frac 8 {11}, 10\frac {10} {11})## with ##Z = 51.818182 \text{ or } Z = $5181.8182##. However, the correct solution requires positive integers, because the solution is about amount of goods (plants), which can't be fractional. How can I solve this problem, and would it become integer programming problem? The correct solution is 3 A plants and 10 S plants, with $5000 optimum value.

Cheers, Bagas

In grade 11 of high school, I encountered this linear programming problem on my textbook:

The "alternative solution" described in the textbook as follows:

Let:

- ##x## : amount of plant A

- ##y## : amount of plant S

- ##L## : garden area

- ##L_x## : area of garden for one plant A

- ##L_y## : area of garden for one plant B

Since the plant order requirement is 20% of total orders are for plant A, we write: $$x \geq \frac 1 5 \left( x + y \right) \text{or } 4x - y \geq 0$$

However, the garden capacity is only for 10 A plants and 15 S plants, that is $$ x \cdot (\frac 1 {10} L) + y \cdot (\frac 1 {15} L \leq L) \text{ or } 3x + 2y \leq 30$$

The objective function is ##Z = 5x + 3.5y## (in the order of $100).

The complete linear problem equations are:

\begin{cases}

4x - y \geq 0 \\

3x + 2y \leq 30 \\

x \geq 0 \\

y \geq 0 \\

Z = 5x + 3.5y \text{ } \text{(maximize)}

\end{cases}

Solving the problem (using graphical method), the solutions are: ##B(2\frac 8 {11}, 10\frac {10} {11})## with ##Z = 51.818182 \text{ or } Z = $5181.8182##. However, the correct solution requires positive integers, because the solution is about amount of goods (plants), which can't be fractional. How can I solve this problem, and would it become integer programming problem? The correct solution is 3 A plants and 10 S plants, with $5000 optimum value.

Cheers, Bagas