5.3 Q-9

Question Statement

We need to maximize the profit function for a machine that can produce two types of products:

• Product A: Requires 2 units of chemical and 1 unit of compound.
• Product B: Requires 1 unit of chemical and 2 units of compound.

The available resources are:

• 800 units of chemical
• 1000 units of compound

The profits per unit of product A and B are Rs. 30 and Rs. 20, respectively.

We need to determine the number of units of each product (A and B) that should be produced to maximize the profit, subject to the constraints on the available resources.

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Background and Explanation

This is a linear programming problem where we need to maximize a profit function while adhering to given constraints. The constraints arise from the limited availability of chemical and compound. The goal is to determine the optimal number of units of products A and B that can be produced without exceeding the available resources, while maximizing the total profit.

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Solution

Let:

• $x$ be the number of units of product A
• $y$ be the number of units of product B

Step 1: Define the Profit Function

The profit function is given by:

$$f(x, y) = 30x + 20y$$

Where:

• $30x$ is the profit from product A
• $20y$ is the profit from product B

Step 2: Write the Constraints

The constraints are based on the available resources:

1. Chemical constraint:

$$2x + y \leq 800$$

2. Compound constraint:

$$x + 2y \leq 1000$$

3. Both $x$ and $y$ must be non-negative:

$$x \geq 0, \quad y \geq 0$$

Step 3: Find the Corner Points

To solve this linear programming problem, we first need to graph the constraints and find the corner points where the feasible region is formed.

1. For the chemical constraint $2x + y = 800$:

• If $x = 0$, then $y = 800$, so the point is $(0, 800)$.
• If $y = 0$, then $x = 400$, so the point is $(400, 0)$.

2. For the compound constraint $x + 2y = 1000$:

• If $x = 0$, then $y = 500$, so the point is $(0, 500)$.
• If $y = 0$, then $x = 1000$, so the point is $(1000, 0)$.

3. Find the intersection of the two lines:

Substitute the expression for $y$ from the chemical constraint into the compound constraint:

$$y = 800 - 2x$$

Substitute into $x + 2y = 1000$:

$$x + 2(800 - 2x) = 1000$$

Simplify and solve for $x$:

$$x + 1600 - 4x = 1000 -3x = -600 x = 200$$

Now substitute $x = 200$ into $y = 800 - 2x$ to find $y$:

$$y = 800 - 2(200) = 400$$

So, the intersection point is $(200, 400)$.

Step 4: Evaluate the Profit at the Corner Points

Now that we have the corner points $(200, 400), (0, 500), (400, 0), (0, 0)$, we can evaluate the profit function at each of these points.

• At $(200, 400)$:

$$f(200, 400) = 30(200) + 20(400) = 6000 + 8000 = 14000$$

• At $(0, 500)$:

$$f(0, 500) = 30(0) + 20(500) = 0 + 10000 = 10000$$

• At $(400, 0)$:

$$f(400, 0) = 30(400) + 20(0) = 12000 + 0 = 12000$$

• At $(0, 0)$:

$$f(0, 0) = 30(0) + 20(0) = 0 + 0 = 0$$

Step 5: Determine the Maximum Profit

The maximum profit occurs at the point $(200, 400)$ with a profit of Rs. 14,000.

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Key Formulas or Methods Used

• Profit Function: $f(x, y) = 30x + 20y$

• Constraints:

  1. $2x + y \leq 800$
  2. $x + 2y \leq 1000$
  3. $x \geq 0, , y \geq 0$

• Linear Programming: Find the intersection points of the constraints and evaluate the profit function at each corner point.

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Summary of Steps

1. Define the profit function: $f(x, y) = 30x + 20y$.
2. Write the constraints based on resource limits.
3. Find the intersection points of the constraints by solving the equations.
4. Evaluate the profit function at each corner point.
5. The corner point $(200, 400)$ gives the maximum profit of Rs. 14,000.

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