Question

class 12 ch 12 Example 8 (Manufacturing problem) A manufacturing company makes two models
A and B of a product. Each piece of Model A requires 9 labour hours for fabricating
and 1 labour hour for finishing. Each piece of Model B requires 12 labour hours for
fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum
labour hours available are 180 and 30 respectively
. The company makes a profit of
Rs 8000 on each piece of model A and Rs 12000 on each piece of Model B. How many
pieces of Model A and Model B should be manufactured per week to realise a maximum
profit? What is the maximum profit per week?​

Answer 1

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

Let us suppose that

➢ Number of pieces of Model A = x

and

➢ Number of pieces of Model B = y

According to statement,

➢ Profit on Model A is Rs 8000 and on Model B is Rs 12000.

➢ Let suppose total profit on selling all pieces of Model A and Model B be Z.

So,

➢ Total profit, Z = 8000x + 12000y.

So, Lets formulate the Linear Programming Problem.

$\rm :\longmapsto\:Maximize \: Z \: = \: 8000x + 12000y$

Subject to the constraints,

$\rm :\longmapsto\:9x + 12y \leqslant 180 - - - (1)$

$\rm :\longmapsto\:x + 3y \leqslant 30 - - - (2)$

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

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

Let consider the first constraint.

$\rm :\longmapsto\:9x + 12y = 180$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:9 \times 0 + 12y = 180$

$\rm :\longmapsto\:12y = 180$

$\bf\implies \:y = 15$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:9x + 12 \times 0 = 180$

$\rm :\longmapsto\:9x = 180$

$\bf\implies \:x = 20$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 15 \\ \\ \sf 20 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 15) & (20 , 0)

➢ See the attachment graph.

Let consider the second constraint.

$\rm :\longmapsto\:x + 3y = 30$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 + 3y = 30$

$\rm :\longmapsto\:3y = 30$

$\rm :\longmapsto\:y = 10$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x + 3 \times 0 = 30$

$\bf\implies \:x = 30$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 10 \\ \\ \sf 30 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 10) & (30 , 0)

➢ See the attachment graph.

Now from graph, we concluded that feasible region is bounded.

So, the value of Z at each corner of feasible region is as follow : -

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Corner \: point & \bf Z = 8000x + 12000y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf (0,0) & \sf 0 \\ \\ \sf (20,0) & \sf 160000\\ \\ \sf (12,6) & \bf 168000\\ \\ \sf (0,10) & \sf 120000 \end{array}} \\ \end{gathered}$

➢ Thus, maximum value of Z is 168000 at (12, 6).

➢ Hence, Company should manufactured 12 pieces of Model A and 6 pieces of Model A to get the maximum Profit.