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solution

Let x packages of nuts and y packages of bolts be produced. The objective of the manufacturer to maximize the profit is

Total Time required on machine 1 to produce x packages of nuts and y packages of bolts is equal to

Total Time required on the machine 2 to produce x packages of nuts and y packages of bolts is equal to

 s

According to restrictions,

For machine 1

For machine 2

Maximize z is equal to

Subject to constraints

To solve this graphically, let us take

The lines are drawn using suitable points on the graph.

The lines intersect at P(3,3)

Now shade the region of intersection of the lines.

The feasible region is OAPB

For the corner point O(0,0), z=

For the corner point A(4,0)

For the corner point P(3,3,)

For the corner point B(0,4)

Clearly z is maximum at x=3 , y=3 and the maximum value is 10.50

2.50x + 1y

1x + 3y

3x + 1y

X + 3y ≤ 12

3x + y ≤ 12 and x,y≥0

2.50x + y

X + 3y ≤ 12

3x + y ≤ 12

X ≥ 0, y ≥ 0

X + 3y =12

3x + y = 12, x=0, y=0

O(0,0),A(4,0),P(3,3,),B(0,4)

2.5(0) +1(0)=0

2.5(4)+1(0)=10

2.5(3)+1(3)=10.5

2.5(0)+1(4)=4

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solution

Let x be the number of wall hangings and y be the number of artificial plants.

Profit of dealer is equal to

Objective function z is equal to

Since dealer invest atmost Rs. 15000

Therefore

Or

Also a dealer has space to store atmost 80 pieces.

Therefore,

Maximize z is equal to

Subject to constraints

To solve this graphically, we need to draw the graph

Let us the draw the lines 2x + y ≤100

X + y ≤ 80

X ≥ 0, y ≥ 0

On the graph by using suitable points.

The points of intersection are

Then shade the region of intersection of these two lines

The feasible points are OABC

Now to obtain the maximum value;

For The corner point O(0,0), z=

For the corner point A(50,0)

For the corner point B(20,60)

For the corner point C(0,100)

Clearly we get the maximum value of 2500 at A.(50,0)

50x + 18y

50x + 18y

300x + 150y ≤ 15000

2x + y ≤ 100

X + y ≤ 80

X ≥ 0, y ≥ 0

Z = 50x + 18y

2x + y ≤100

X + y ≤ 80

X ≥ 0, y ≥ 0

(20,60)

O(0,0),A(50,0),B(20,60),(C(0,100)

50(0)+18(0)=0

50(50)+18(0)=2500

50(20)+18(60)=2080

50(0)+18(100)=1800