Question

A company manufactures two types of sweaters: type A and type B it costs Rs. $ 360 $ to make a type A sweater and Rs. $ 120 $ to make a type B sweater. The company can make at most $ 300 $ sweaters and spend at most Rs. $ 72000 $ a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than $ 100 $ . The company makes a profit of Rs. $ 200 $ for each sweater of type $ A $ and $ Rs.120 $ for every sweater of type $ B $ . Formulate this problem as a $ LPP $ to maximize the profit to the company $ 15 $ . A man rides his motorcycle at the speed of $ 50 $ km/hour. He has to spend

Answer 1

Hint: Firstly, we will suppose the number of sweaters of type A and type B. Further we will convert the statement into an equation. Then, we will plot the points on the graph. Thereafter, the point will be given the maximum value to get the result.

Complete step-by-step answer:
Let the number of sweater type A $ = x $
Let the number of sweater type B $ = y $
The company spends at most Rs. $ 72000 $ a day.
According to the question $ 360x + 120y \leqslant 72000 $
We will take $ 120 $ common, here
 $ 120(3x + y) \leqslant 72000 $
 $ 3x + y \leqslant \dfrac{{72000}}{{120}} $
 $ 3x + y \leqslant 600 $ ……(i)
And, company can make at most $ 300 $ sweaters
 $ \therefore x + y \leqslant 300 $ ……(ii)
Also, the number of sweaters of type B cannot exceed
The number of sweater of type A are more than $ 100 $
 $ \therefore x - y \leqslant 100 $ ……(iii)
Now, the company makes a profit of Rs. $ 200 $ for each sweater of type A and Rs. $ 120 $ for every sweater of type B.
Therefore $ z = 200x + 120y $ …..(iv)
 $ x + y \leqslant 300 $
If $ x = 0, $ then $ y = 300 $
If $ y = 0 $ then $ x = 300 $

$ x $	$ 0 $	$ 300 $
$ y $	$ 300 $	$ 0 $

 Now, $ x - y = 100 $
If $ x = 0,\,y = - 100 $
If $ y = 0,x = 100 $

$ x $	$ 0 $	$ 100 $
$ y $	$ - 100 $	$ 0 $

 $ 3x + y \leqslant 600 $
If $ x = 0,\,\,y = 600 $
If $ y = 0,\,\,x = 200 $

$ x $	$ 0 $	$ 200 $
$ y $	$ 600 $	$ 0 $

Getting vertices of flexible regions, $ A(0,300) $
 $ G(150,150),\,H(200,100),\,I(175 - 75) $
We can put these values in equation (iv), we have
 $ G(150,150) = 200 \times 150 + 120 \times 150 $
 $ = 30000 + 18000 $
 $ = 48000 $
 $ I(175,75) = 200 \times 175 + 120 \times 75 $
 $ = 35000 + 9000 = 44,000 $
 $ A(0,300) = 200 \times 0 + 120 \times 300 $
 $ = 0 + 36000 $
 $ = 36000 $
 $ D(100,0) = 200 \times 100 + 120 \times 0 $
 $
  20,000 + 0 \\
   = 20,000 \\
  $
 $ H(200,100) = 200 \times 100 + 120 \times 100 $
 $
   = 20,000 + 12000 \\
   = 32000 \\
  $
Maximum profit(P) $ = 200(175) + 120 \times 75 $
 $ (P) = Rs.35000 + 9000 $
 $ (P) = Rs.44000 $
So, type A\[ = 175\,\,and\,\,type\,\,B = 75\]

Note: Students should plot the points carefully on the graph and the maximum value of that point will give the highest values in your answer.