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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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A manufacturing company makes two types of teaching aids A and B of Mathematics for class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for finishing. Each type of B requires 12 labour hours of fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available per week are 180 and 30 respectively. The company makes a profit of Rs. 80 on each piece of type A and Rs. 120 on each piece of type B. How many pieces of type A and B should be manufactured per week to get a maximum profit? Make it as an LPP and solve graphically. What is the maximum profit per week?

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Let the pieces of type A manufactured per week be x
Let the pieces of type B manufactured per week be y
$ \therefore $ maximum profit Z = $80 x + 120 y$
Fabricating hours for A is 9 and finishing hours is 1
Fabricating hours for B is 12 and finishing hours is 3
Maximum number of fabricating hours = 180
$ \therefore 9x + 12y \leq 180$
i.e., $ 3x+4y \leq 60$
Maximum number of finishing hours = 30
$ \therefore x + 3y \leq 30$
Now Z is $ = 80x+120y$ is subjected to
Constraints $ 3x+4y \leq 60$
$ x + 3y \leq 30$
$x \geq 0$
$y \geq 0$
Now the area of the fesible region is as shown in the figure
$3x+4y=60 \quad \quad \therefore y = 6$
$(3)x+3y=30 \quad \quad and \: x = 12$
$ 3x+4y=60$
$ 3x+9y=30$
We can see that the points on the bounded region are
A(0, 15), B( 0, 10), C(20, 0) and D(12, 6)
Now let us calculate the maximum profit
$Points (x,y) \quad \quad \quad Z = 80x + 120y$
$A(0, 15) \quad \quad \quad \quad \quad Z = 80(0) + 120(15) = 1800$
$B(0, 10) \quad \quad \quad \quad \quad Z = 80(0) + 120(10) = 1200$
$C(20, 0) \quad \quad \quad \quad \quad Z = 80(20) + 120(0) = 1600$
$D(12, 6) \quad \quad \quad \quad \quad Z = 80(12) + 120(6) = 960+720 = 1680$
Hence the maximum profit is at (0, 15)
$ \therefore $ teaching aid A = 0 and teaching aid B = 15 has to be made.
answered Apr 1, 2014 by thanvigandhi_1
 

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