Test: Linear Programming (CBSE Level) - 2 - Commerce MCQ

Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,then , optimal value must occur at a corner point (vertex) of the feasible region.

Objective function is Z = 3x + 4 y ……(1).
The given constraints are : x + y ≤ 4, x ≥ 0, y ≥ 0.

therefore Z = 16 is maximum at ( 0 , 4 ) .

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.

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

Here , maximize Z = 11x + 7y , subject to the constraints :2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0.

Let R be the feasible region for a linear programming problem, and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R .

Objective function is Z = - 3x + 4 y ……………………(1).
The given constraints are : x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Here , Z = -12 is minimum at C (4 , 0) .

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 .

Here , maximize , Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0.

Let R be the feasible region for a linear programming problem,and let Z = ax + by be the objective function. If R is bounded, then the objective function Z has both a maximum and a minimum value on R and each of these occurs at a corner point (vertex) of R.

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

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.

We have , Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2,x ≥ 0, y ≥ 0.

Corner Points(4,0)(2,1)(0,3)​Corresponding value of Z 3 (Minimum)​

From the shaded region, it is clear that feasible region is unbounded with the corner points A(4, 0, B(2, 1) and C(0, 3).

Also, we have Z=4x+y ,

Now, we see that 3 is the smallest value of Z at the corner point (0, 3). Note that here we see that the region is unbounded, therefore 3 may or may not be the minimum value of Z.

To decide this issue, we graph the inequality 4x+y<3 and check whether the resulting open half plan has no point in common with feasible region otherwise, Z has no minimum value.

From the shown graph above, it is clear that there is no point in common with feasible region and hence Z is minimum value of 3 at (0, 3).

A maximum or a minimum may not exist for a linear programming problem if The feasible region is unbounded .

Objective function is Z = 3x + 2 y ……………………(1).
The given constraints are : x + 2y ≤ 10, 3x + y ≤ 15, x, y ≥ 0.

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 .

In Corner point method for solving a linear programming problem the first step is : To find the feasible region of the linear programming problem and determine its corner points (vertices)

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

Here , Z = 18 is minimum at (0, 3) and (6 , 0) .
Minimum Z = 6 at all the points on the line segment joining the points (6, 0) and (0, 3).

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.

We have , Maximize Z = 100x + 120y , subject to constraints 2x + 3y ≤ 30, 3x + y ≤ 17, x ≥ 0, y ≥ 0.