# NCERT Class 12-Mathematics: Chapter – 12 Linear Programming Part 4 (For CBSE, ICSE, IAS, NET, NRA 2022)

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## Fill in the Blanks in Each of the Examples 9 and 10

**Question 9**:

In a LPP, the linear function which has to be maximised or minimised is called a linear ________ function.

**Answer**:

Objective

**Question 10**:

The common region determined by all the linear constraints of a LPP is called the ________ region.

**Answer**:

Feasible.

### State Whether the Statements in Examples 11 and 12 Are True or False

**Question 11**:

If the feasible region for a linear programming problem is bounded, then the objective function has both a maximum and a minimum value on R.

**Answer**: True

**Question 12**:

The minimum value of the objective function in a linear programming problem always occurs at only one corner point of the feasible region.

**Answer**: False

The minimum value can also occur at more than one corner points of the feasible region.

## 12.3 EXERCISE

### Short Answer (S. A)

**Question 1**:

Determine the maximum value of subject to the constraints: .

**Answer**:

We have, maximise

Subject to the constraints

We see that, the feasible region as shaded determined by the system of constraint (ii) to (iv) is OABC and bounded. So, now we shall use corner point method to determine the maximum value of .

Corner Points | Corresponding Value of Z |

Hence, the maximum value of Z is 42 at

**Question 2**:

Maximise Z = 3*x* + 4*y*, subject to the constraints: .

**Answer**:

Given:

It is subject to constraints

Now let us convert the given inequalities into equation.

We obtain the following equation

The region represented by :

The line meets the coordinate axes and respectively. We will join these points to obtain the line . It is clear that satisfies the inequation . So the region containing the origin represents the solution set of the in equation

Region represented by and is first quadrant, since every point in the first quadrant satisfies these in equations.

Plotting these equations graphically, we get

The shaded region OBC shows the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.

Corner Points are and .

Now we will substitute these values in Z at each of these corner points, we get

Corner Point | Value of |

Hence, the maximum value of Z is at the point .