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  • A company makes 3 model of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculator of model B and 4800 calculator of model C. At factory I, 50 calculators of model A, 50 of mode

A company makes 3 model of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculator of model B and 4800 calculator of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made every day. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.

Answers (1)

Taking into consideration that Let number of days for which factory I operate be x and number of days for which factory II operates be y.

Number of calculators made by factory I and II of model A are 50 and 40 respectively.

Minimum number of calculators of model A required = 6400

So, 50x + 40y ≥ 6400

⇒ 5x + 4y ≥ 640

Number of calculators made by factory I and II of model B are 50 and 20 respectively.

Minimum number of calculators of model B required = 4000

So, 50x + 20y ≥ 4000

⇒ 5x + 2y ≥ 400

Number of calculators made by factory I and II of model C are 30 and 40 respectively.

Minimum number of calculators of model C requires = 4800

So, 30x + 40y \geq 4800

\Rightarrow 3x + 4y \geq 480

Operating costs is Rs 12000 and Rs 15000 each day to operate factory I and II respectively.

Let Z be total operating cost so we have Z = 12000x + 15000y

Also, number of days are non-negative so, x, y ≥ 0

So, we have,

Constraints,

\\ 5x + 4y $ \geq $ 640\\ \\ 5x + 2y $ \geq $ 400\\ \\ 3x + 4y $ \geq $ 480\\ \\ x, y $ \geq $ 0\\ \\ Z = 12000x + 15000y\\

We need to minimize Z, subject to the given constraints.

Now let us convert the given inequalities into equation.

We obtain the following equation

\\ 5x + 4y $ \geq $ 640\\ \\ $ \Rightarrow $ 5x + 4y = 640\\ \\ 5x + 2y $ \geq $ 400\\ \\ $ \Rightarrow $ 5x + 2y = 400\\ \\ 3x + 4y $ \geq $ 480\\ \\ $ \Rightarrow $ 3x + 4y = 480\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow $ y=0\\

The region that represents the 5x + 4y ≥ 640.

The line 5x + 4y = 640 meets the coordinate axes (128,0) and (0,160) respectively to get the final outcome. When we join these points to obtain the line 5x + 4y = 640. It is then clear that (0,0) does not satisfy the inequation 5x + 4y ≥ 640. So, the region not containing the origin represents the solution set of the inequation 5x + 4y ≥ 640.

The region that represents the 5x + 2y ≥ 400:

The line 5x + 2y = 400 meets the coordinate axes (80,0) and (0,200) respectively. We will join these points to obtain the line 5x + 2y = 400. It is justified that (0,0) does not satisfy the inequation 5x + 2y ≥ 400. So, the region not containing the origin represents the solution set of the inequation 5x + 2y ≥ 400.

The region represented by 3x + 4y ≥ 480:

The line that 3x + 4y = 480 when meets the coordinate axes (160,0) and (0,120) respectively. We will then try and join these points to obtain the line 3x + 4y = 480. It is clear that (0,0) does not justify the inequation 3x + 4y ≥ 480. So, the region not containing the origin represents the solution set of the inequation 3x + 4y ≥ 480.

The graph is:

The region towards the right of ABCD is the feasible region. It is unbounded in this case.

The value of Z at the corner points, is

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=12000 \times+15000 y \\ \hline A(0,200) & Z=3000000 \\ \hline B(32,120) & Z=218400 \\ \hline C(80,60) & Z=1860000 \rightarrow \text { min } \\ \hline D(160,0) & Z=1920000 \\ \hline \end{array}

Now, we plot 12000 x+15000 y<1860000 to check if resulting open half has any point common with feasible region.
The region represented by 12000 \mathrm{x}+15000 \mathrm{y}<1860000
The line 12000 \mathrm{x}+15000 \mathrm{y}=1860000 meets the coordinate axes (155,0) and (0,124) respectively. We will join these points to obtain the line $12000 \mathrm{x}+15000 \mathrm{y}=1860000. It is clear that (0,0) satisfies the inequation 12000 \mathrm{x}+15000 \mathrm{y}<1860000. So, the region containing the origin represents the solution set of the inequation 12000 \mathrm{x}+15000 \mathrm{y}<1860000.

Clearly, 12000 \mathrm{x}+15000 \mathrm{y}<1860000 intersects feasible region only at C
So, value of Z is minimum at  C(80,60), the minimum value is 1860000 .
So, number of days factory 1 is required to operate is 80 and number of days factory 2 should operate is 60 to minimize the cost.

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