Question

A manufacturing company produces chairs and tables. A chair requires 3 hours on machine ‘A’ and 2 hours on machine ‘B’, while a table requires 2 hours on machine ‘A’ and 4 hours on machine ‘B’. The machine can operate for only 8 hours a day. The profit on a chair is Rs 420, and the profit on each table is Rs 360. (a) find how many of each product should be produced to maximize the profit? [06] (b) what will be the amount of optimum profit?​

Answer 1

Answer:

Let daily production of chairs and tables be x and y respectively.

Since, profits of each chair and table is Rs. 3 and Rs. 5 respectively. So, profits on x number of type A and y number of type B are 3x and 5y respectively.

Let Z denotes total output daily, so,

Z = 3x + 5y

Since, each chair and table requires 2 hrs and 3 hrs on machine A respectively. So, x number of chair and y number of table require 2x and 4y hrs on machine A respectively. But,

Total time available on Machine A is 16 hours. So,

2x + 3y

16

x + 2y

8 {First Constraint}

Since, each chair and table requires 6 hrs and 2 hrs on machine B respectively. So, x number of chair and y number of table require 6x and 2y hrs on machine B respectively. But,

Total time available on Machine B is 30 hours. So,

6x + 2y

30

3x + y

15 {Second Constraint}

Hence mathematical formulation of the given LPP is,

Max Z = 3x + 5y

Subject to constraints,

x + 2y

8

3x + y

15

x,y

0 [Since production of chairs and tables can not be less than zero]

Region x + 2y

8: line x + 2y = 8 meets the axes at A(8,0), B(0,4) respectively.

Region containing the origin represents x + 2y

8

as origin satisfies x + 2y

8.

Region 3x + y

15: line 3x + y = 15 meets the axes at C(5,0), D(0,15) respectively.

Region containing the origin represents 3x + y

15 as origin satisfies 3x + y

15

Region x,y

0: it represents the first quadrant.

10.jpg

The corner points are O(0,0), B(0,4), E(

), and C(5,0).

The values of Z at these corner points are as follows,

The maximum value of Z is 22.2 which is attained at E(

).

Thus the maximum profit of Rs 22.2 when

units of chair and

units of table are produced.

I hope it will helps you.

Please mark me as brainliest.

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that

Production of chairs be x units

and

Production of tables be y units.

According to statement

Profit on chair is Rs. 420

Profit om table is Rs 360.

So, profits on x number of chairs be Rs 420x

and

Profit on y number of tables be Rs 360y.

Let assume that Z denotes total Profit.

So, Z = 420x + 360y

According to statement,

Each chair and table requires 3 hrs and 2 hrs on machine A respectively.

So, x number of chair and y number of table require 3x and 3y hrs on machine A respectively. But,

Total time available on Machine A is 8 hours. So

$\rm :\longmapsto\:3x + 2y \leqslant 8 - - - (1)$

According to statement,

Since, each chair and table requires 2 hrs and 4 hrs on machine B respectively.

So, x number of chair and y number of table require 2x and 4y hrs on machine B respectively. But,

Total time available on Machine B is 8 hours. So,

$\rm :\longmapsto\:2x + 4y \leqslant 8$

$\rm :\longmapsto\:x + 2y \leqslant 4 - - - (2)$

So, the required LPP is

$\red{\rm :\longmapsto\:Maximize \: Z \: = 420x + 360y}$

Subject to the constraints,

$\:\rm :\longmapsto\:3x + 2y \leqslant 8$

$\:\rm :\longmapsto\:x + 2y \leqslant 4$

$\rm :\longmapsto\:x,y \: \geqslant 0$

Let find the feasible region by plotting the graph.

Consider,

$\rm :\longmapsto\:3x + 2y = 8$

On substituting x = 0, we get

$\rm :\longmapsto\:0 + 2y = 8$

$\rm :\longmapsto\:y = 4$

On substituting y = 0, we get

$\rm :\longmapsto\:3x + 0 = 8$

$\rm :\longmapsto\:x = \dfrac{8}{3}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf \dfrac{8}{3} & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

Consider,

$\rm :\longmapsto\:x + 2y = 4$

On substituting x = 0, we get

$\rm :\longmapsto\:0 + 2y = 4$

$\rm :\longmapsto\:y = 2$

On substituting y = 0, we get

$\rm :\longmapsto\:x + 0 = 4$

$\rm :\longmapsto\:x = 4$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 2 \\ \\ \sf 4 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

Now, from graph we concluded that OABC is a feasible closed region.

So, corner points to find the maximum profit are

A(8/3, 0), B (2,1) and C(0,2)

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Corner \: Points & \bf Z = 420x + 360y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \bigg(\dfrac{8}{3},0 \bigg) & \sf 1120 \\ \\ \sf (2,1) & \sf 1200\\ \\ \sf (0,2) & \sf 720 \end{array}} \\ \end{gathered}$

So, we find that the maximum profit is Rs 1200 when 2 chairs and one table is sold.

Hence,

Maximum profit = Rs 1200

Number of chair produced = 2

Number of table produced = 1