Linear Programming - Online Test

Objective function is Z = - x + 2 y ……………………(1).
The given constraints are : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

Here , the open half plane has points in common with the feasible region .
Therefore , Z has no maximum value.

Let number of cakes of first type = x
And number of cakes of second type = y
Therefore , the above L.P.P. is given as :
Minimise , Z = x +y , subject to the constraints : 200x +100y ≤ 5000 and. 25x +50y ≤ 1000, i.e. 2x + y ≤ 50 and x +2y ≤ 40 x, y ≥ 0.

The corner points can be obtained by constructing the lines x+2y=40 , 2x+y= 50 and x+2y = 40.

The points so obtained are (0,0),(25,0), (20,10), and (0,20).

Here Z = 30 is maximum.
i.e Maximum number of cakes = 30 , 20 of kind one and 10 cakes of another kind .

Let number of rackets made = x
And number of bats made = y
Therefore , the above L.P.P. is given as :
Maximise , Z = x +y , subject to the constraints : 1.5x +3y ≤ 42 and. 3x +y ≤ 24, i.e.0.5x + y ≤ 14 i.e. x +2y ≤ 28 and 3x +y ≤ 24 , x, y ≥ 0.

Here Z = 16 is maximum. i.e Maximum number of rackets = 4 and number of bats = 12.

Let number of rackets made = x
And number of bats made = y
Therefore , the above L.P.P. is given as :
Maximise , Z = x +y , subject to the constraints : 1.5x +3y ≤ 42 and. 3x +y ≤ 24, i.e.0.5x + y ≤ 14 i.e. x +2y ≤ 28 and 3x +y ≤ 24 , x, y ≥ 0.

Here Z = 16 is maximum. i.e Maximum number of rackets = 4 and number of bats = 12.
Here , profit function is P = 20x + 10y
Profit is maximum at x = 4 and y = 12 .
Therefore , maximum profit = 20(4) + 10 ( 12) = 200.i.e. Rs.200.

Let number of packages of nuts produced = x
And number of packages of bolts produced = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 17.50x +7y , subject to the constraints : x +3y ≤ 12 and. 3x +y ≤ 12, x, y ≥ 0.

Here Z = 73.50 is maximum.
i.e 3 packages of nuts and 3 packages of bolts;
Maximum profit = Rs 73.50.

Let number of packages of screws A produced = x
And number of packages of screws B produced = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 7x +10y , subject to the constraints : 4x +6y ≤ 240 and. 6x +3y ≤ 240 i.e. 2x +3y ≤ 120 and 2x +y ≤ 80 , x, y ≥ 0.

Here Z = 410 is maximum.
i.e 30 packages of screws A and 20 packages of screws B; Maximum profit = Rs 410.

Let number of pedestal lamps manufactured = x
And number of wooden shades manufactured = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 5x +3y , subject to the constraints : 2x +y ≤ 12 and. 3x +2y ≤ 20 , x, y ≥ 0.

Here Z = 32 is maximum.
i.e 30 packages of screws A and 20 packages of screws B; Maximum profit = Rs 410.
i.e. 4 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 32 .

Let number of souvenirs of type A = x
And number of souvenirs of type B = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 5x +6y , subject to the constraints : 5x +8y ≤ 200 and. 10x +8y ≤ 240 , x, y ≥ 0.

Here Z = 160 is maximum.
i.e. 8 Souvenir of types A and 20 of Souvenir of type B; Maximum profit = Rs 160.

Let number of desktop model computers = x
And number of portable model computers = y
Therefore , the above L.P.P. is given as :
Maximise , Z = 4500x +5000y , subject to the constraints : x +y ≤ 250 and 25000x +40000y ≤ 700000 .i.e. x +y ≤ 250 and 5x +8y ≤ 1400 , x, y ≥ 0.

Here Z = 1150000 is maximum.
i.e. 200 units of desktop model and 50 units of portable model; Maximum profit = Rs 1150000 .

Let number of units of food F1 = x
And number of units of food F2 = y
Therefore , the above L.P.P. is given as :
Minimise , Z = 4x +6y , subject to the constraints : 3 x + 6y ≥ 80, 4x + 3y ≥ 100, x,y ≥ 0.,

Corner points Z =4x +6 y B(80/3 , 0 ) 320/3 D(24,4/3 ) 104…………………(Min.) A(0,100/3) 200 Here Z = 104 is minimum. i.e. Minimum cost = Rs 104.