Maths M.C.Q (1 Marks)

2.  $\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ given an optimal solution

Solution:

A set A is convex if, for any two points X₁, X₂ $\in\text{A}$ and $\lambda\in0,1$ imply that $\lambda\times1+1-\lambda\times2\in\text{A}$.

Since, here X₁ and X₂ are optimal solution

Therefore, their convex combination will also be an optimal solution

Thus, $\text{X}=\lambda\ \text{X}_1+(1-\lambda)\text{X}_2,0\leq\lambda\leq1$ gives an optimal solution.

3. I and III

Solution:

To write the dual, then all the primal variables must be non-negative.

All the constraints are $\leq$ type if it ia maximization problem and $\geq$ type if it is a minimization problem.

4. None of the above

Solution:

A constraint in an LP model becomes redundant when the feasible region doesnt change by the removing the constraint.

For example, $\text{x}+2\text{y}\leq20$ and $2\text{x}+4\text{y}\leq40$ are the constraints.

$2\text{x}+4\text{y}\leq40$

$\Rightarrow2\times(\text{x}+2\text{y})\leq2\times20$

$\Rightarrow\text{x}+2\text{y}\leq20$

 which is same as the first constraint.

Therefore, $2\text{x}+4\text{y}\leq40$ can be removed.

By removing this constraint feasible region doesnt change.

2. Non - negativity restriction.

Solution:

The feasible region is the set of all the points that satisfy all the given constraints.

The variables of the linear programs must always take the non - negative values (i.e., $\text{x}\geq0$ and $\text{y}\geq0$).

These are used because x and y are usually the number of items produced and we cannot produce the negative number of items.

The least possible number of items could be zero.

Therefore, the feasible solution should satisfy the non - negativity restriction.

4. Has no feasible solution

Solution:

$\text{x}+2\text{y}\leq2$

$\text{x}+2\text{y}\geq8$

$\text{x},\text{y}\geq0.$

4. $\{(\text{x},\text{y}):\text{y}\geq2,\text{y}\leq4\}$

Solution:

is the region between two parallel lines, so any line segment joining any two points in it lies in it.

Hence, it is a convex set.

1. 59 at$\Big(\frac{9}{2},\frac{5}{2}\Big)$

4. 0.8645

Solution:

Let A = Part X is not defective

Probability of A is $\text{P}(\text{A})=\frac{91}{100}$

B = Part Y is not defective.

Probability of B is $\text{P}(\text{B})=\frac{95}{100}$

Required probability

 $=\text{P}(\text{A}\cap\text{B})=\text{P}(\text{A})\text{P}(\text{B})=\frac{91}{100}\times\frac{95}{100}=\frac{8645}{10000}$

3. $\Big(2,\frac{7}{2}\Big)$

2. Two subsets

Solution:

In graphical solutions of linear inequalities, solution can be divided into two subsets.

for example, $2\text{x}+\text{y}\leq4$

One subset includes all values (x, y) that satisfy the equation 2x + y = 4 and the other subset includes all the values (x, y) that satisfy 2x + y < 4.

1. 8

Solution:

First, rewrite each equation, so that it is in the slope - intercept form of a line, which is y = mx + b, where mm is the slope and b is the y - intercept of the line.

The first equation becomes 2y < x + 4 or $\text{y}\leq\frac{1}{2}\text{x}+2.$

The second equation becomes $\text{y}\geq2\text{x}-4.$

The greatest x + y is the point at which the two lines intersect.

Set the equations of the two lines, $\text{y}=\frac{1}{2}\text{x}+2$ and y = 2x - 4, equal to each other and solve for x.

The resulting equation is $\text{y}=\frac{1}{2}\text{x}+2$ 

and y = 2x - 4.

Solve for x to get $\text{y}=\frac{3}{-2}\text{x}+2=-4$ or $\frac{3}{-2}\text{x}=-6,$

$\Rightarrow\text{x}=4$

Next, plug 4 into one of the two equations to solve for y.

Therefore, y = 2(4) - 4 = 4 and x + y = 4 + 4 = 8.

4. $(0,5)$