CBSE Class 12 Math 2020 Outside Delhi Set 1 Solved Paper

Question : 31 of 36

Marks: +1, -0

A manufacturer has three machines I, II and III installed in his factory. Machine I and II are capable of being operated for atmost 12 hours whereas machine III must be operated for atleast 5 hours a day. He produces only two items

{\M}

and

{\N}

each requiring the use of all the three machines.
The number of hours required for producing 1 unit of

{\M}

and

{\N}

on three machines are given in the following table:

Items sd	Number of hours required on machines
Items sd	I	II	III
${\M}$	1	2	1
${\N}$	2	1	1.25

He makes a profit of

₹ 600

and

₹ 400

on one unit of items

{\M}

and

{\N}

respectively. How many units of each item should he produce so as to maximize his profit assuming that he can sell all the items that he produced. What will be the maximum profit?

Solution:

Let

{\x}

and

{\y}

be the number of items

{\M}

and

{\N}

respectively.
Total profit on the production

={\Rs}(600 {\x}+400 {\y})

Mathematical formulation of the given problem is as follows :
Maximise

Z=600 x+400 y

subject to the constraints :

x+2 y ≤ 12 \;\text " (constraint on Machine I) ...(1) "\;

2 x+y ≤ 12 \;\text " (constraint on Machine II) ...(2) "\;

x+54 y ≥ 5 \;\text " (constraint on Machine III) ...(3) "\;

x ≥ 0, y ≥ 0 . . .(4)

Let us draw the graph of constraints (1) to (4).

{\ABCDE}

is the feasible region (shaded) as shown in Fig determined by the constraints (1) to (4). Observe that the feasible region is bounded, coordinates of the corner points A, B, C, D and E are

(5,0)(6,0),(4,4),(0,6)

and

(0,4)

respectively.
Let us evaluate

Z=600 x+400 y

at these corner points.

Corner point	$Z=600 x+400 y$
$(5,0)$	3000
$(6,0)$	3600
$(4,4)$	$4000 ←$ Maximum
$(0,6)$	2400
$(0,4)$	1600

We see that the point

(4,4)

is giving the maximum value of

Z

. Hence, the manufacturer has to produce 4 units of each item to get the maximum profit of Rs. 4000 .

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