14.
An automobile company manufactures two cars A and B. Model A requires 15 man –
hours for assembly; 5 man - hours for painting and finishing and 1 man hours for
checking and testing. Model B requires 6 man – hours for assembly; 4 man - hours for
painting and finishing and 2 man – hours for checking and testing. There are 400 man -
hours in assembly shop; 150 man - hours in painting and finishing shop ad 40 man –
hours are available in checking and testing division. Express this using linear inequalities :
(a)
(15x + 6y < 400
5x+4y = 150
x + 2 y = 40
x>0; y 20
(b)
(15x + 6y 2 400
5x+4y 2 150
x + 2 y > 40
x20; y 20
(c)
(6x +15y 400
6x + 5y = 150
x + 2 y = 40
x20;y 20
(d)
None of these
Answers
Answer:
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Answer:
ello mate here's your answer! ✋
Step-by-step explanation:
Fabricating Hours Finishing Hours
A 9 1
B 12 3
Let pieces of type A manufactured per week =x
Let pieces of type B manufactured per week =y
Companies profit function which is to be maximized: Z =80x+120y as shown in the tabular column:
Constraints: Maximum number of fabricating hours =180
Therefore, 9x+12y≤180⇒3x+4y≤60
Where 9x is the fabricating hours spent by type A teacher aids, and 12y hours spent on type B and maximum number of finishing hours =30.
x+3y≤30
Where x is the number of hours spent on finishing aid A while 3y on aid B.
So, the LPP becomes:
Z(maximise)=80x+120y
Subject to 3x+4y≤60
x+3y≤30
x≥0
y≥0
Solving it graphically:
Z=80x+120y at (0,15)=1800
Z =1200 at (0,10)
Z =1600 at (20,0)
Z =960+720 at (12,6)=1680
Maximum profit is at (0,15).
Therefore, Teacher aid A =0
Teacher aid B =15 should be made.
/The answer will come itself once u understood the formula thankies/