A company manufactures 2 types of goods P and Q that requires copper and brass. Each unit of type P requires 2 grams of brass and 1 gram of copper while one unit of type Q requires 1 gram of brass an - Mathematics and Statistics

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A company manufactures 2 types of goods P and Q that requires copper and brass. Each unit of type P requires 2 grams of brass and 1 gram of copper while one unit of type Q requires 1 gram of brass and 2 grams of copper. The company has only 90 grams of brass and 80 grams of copper. Each unit of types P and Q brings profit of ₹ 400 and ₹ 500 respectively. Find the number of units of each type the company should produce to maximize its profit

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Solution

Let the company manufacture 'x' units of type P and 'y' units of type Q.

Since number of goods cannot be negative, x ≥ 0, y ≥ 0

Each unit of types P and Q brings profit of ₹ 400 and ₹ 500 respectively.

∴ Total profit = Z = 400x + 500y

We construct a table with the constraints of Brass and Copper as follows:

Material/Type P Q Maximum
Availability
Brass 2 1 90
Copper 1 2 80

From the table, the constraints are

2x + y ≤ 90

x + 2y ≤ 80

∴  Given problem can be formulated as follows:

Maximize Z = 400x + 500y

Subject to 2x + y ≤ 90

x + 2y ≤ 80, x ≥ 0, y ≥ 0

To draw the feasible region, construct table as follows:

Inequality 2x + y ≤ 90 x + 2y ≤ 80
Corresponding equation (of line) 2x + y = 90 x + 2y = 80
Intersection of line with X-axis (45, 0) (80, 0)
Intersection of line with Y-axis (0, 90) (0, 40)
Region Origin side Origin side

Shaded portion OABC is the feasible region, whose vertices are O(0, 0), A(45, 0), B and C(0, 40).

B is the point of intersection of the lines 2x + y = 90 and x + 2y = 80. Solving the above equations, we get

x = `100/3`, y  `70/3`

∴ B ≡ `(100/3, 70/3)`

Here, the objective function is

Z = 400x + 500y

∴ Z at O(0,0) = 400(0) + 500(0) = 0

Z at A(45, 0) = 400(45) + 500(0) = 18000

Z at B`(100/3, 70/3) = 400(100/3) + 500(70/3)`

= `40000/3 + 35000/3`

= `75000/3`

= 25000

Z at C(0, 40) = 400(0) + 500(40)

= 20000

∴ Z has maximum value 25000 at x = `100/3` and y  `70/3`

∴ The company should produce `100/3` units of type P and `70/3` units of type Q to get maximum profit of ₹ 25000.

  Is there an error in this question or solution?
Chapter 2.6: Linear Programming - Q.4 (D)

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