A company produces two different products. One of them needs 1/4 of an hour of assembly work per unit, 1/8 of an hour in quality control work and Rs1.2 in raw materials. The other product requires 1/3 of an hour of assembly work per unit, 1/3 of an hour in quality control work and Rs 0.9 in raw materials. Given the current availability of staff in the company, each day there is at most a total of 90 hours available for assembly and 80 hours for quality control. The first product described has a market value (sale price) of Rs 9 per unit and the second product described has a market value (sale price) of Rs 8 per unit. In addition, the maximum amount of daily sales for the first product is estimated to be 200 units, without there being a maximum limit of daily sales for the second product. Formulate and solve graphically the LPP and find the maximum profit.

Let “a” be the number of units of product 1 produced in one day and b be the number of units of product 2 produced in one day

The selling price of product 1 is 9rs and cost price is 1.2rs hence a profit of 9 – 1.2 = 7.8rs. Profit behind ‘a’ product will be 7.8a

The selling price of product 2 is 8rs and cost price is 0.9rs hence a profit of 8 – 0.9 = 7.1rs. Profit behind ‘b’ products will be 7.1b

Hence total profit daily z = 7.8a + 7.1b

We have to maximize this profit ‘z = 7.8a + 7.1b’ based on some constraints

Let us identify the constraints now

Constraint1: Assembly work

Product 1 requires 1/4 of an hour of assembly work per unit hence for ‘a’ units of an hour of time will be required

Product 2 requires of an hour of assembly work per unit hence for ‘b’ units of an hour of time will be required

Number of hours available for assembly work is 90

Hence the total time of assembly work for product 1 and product 2 should not be greater than 90

Multiply by 12

⇒ 3a + 4b ≤ 1080 …(i)

Constraint2: Quality control work

Product 1 requires of an hour of quality control work per unit hence for ‘a’ units of an hour of time will be required

Product 2 requires of an hour of quality control work per unit hence for ‘b’ units of an hour of time will be required

Number of hours available for quality control work is 80

Hence the total time of quality control work for product 1 and product 2 should not be greater than 80

Multiply by 24

⇒ 3a + 8b ≤ 1920 …(ii)

Constraint3: The maximum amount of sale of product 1 daily is 200 units.

⇒ a ≤ 200 …(iii)

Also, as “a” and “b” represent number of units produced hence it cannot be negative hence a ≥ 0 and b ≥ 0

Plot equations (i), (ii) and (iii) and mark their intersection points

Now in (i) and (ii) less than means below the lines and in (iii) a < 200 means to the left of a = 200

Take scale

On X-axis 1cm = 50 units

On Y-axis 1 cm = 50 units

Now the corner points are F, D, E, G and O

Let us find values of z at these points

Hence the maximum value of z is 2412 at E (200, 120)

Hence the maximum profit is Rs 2412.