Linear Programming Test - 32

Given objective function is \(Z=-50 x+20 y\).

We have to minimize \(Z\) on given constraints

\(2 x-y \geq-5 \)

\(3 x+y \geq 3 \)

\(2 x-3 y \leq 12 \)

\(x \geq 0, y \geq 0\)

After plotting all the constraints we get the common region (Feasible region) as shown in the image.

There are four corner points \((0,5),(0,3),(1,0)\) and \((6,0)\)

Now, at corner points value of \(Z\) are as follows:

\(\begin{array}{|l|l|} \hline Corner~ Points & Value ~of ~Z=-50 x+20 y \\ \hline (0,5) & 100 \\ \hline (0,3) & 60 \\ \hline (1,0) & -50 \\ \hline (6,0) & -300 (minimum) \\ \hline \end{array}\)

Since common region is unbounded. So, value of \(Z\) may be minimum at \((6,0)\) and minimum value may be \(-300\).

Now to check if this minimum is correct or not, we have to draw region \(-50 x+20 y \leq-300\)

Since, there are some common region with feasible region(See image). So, \(-300\) will not be minimum value of \(Z\).

So, \(Z\) has no minimum value.

Given:

Maximize \(\mathrm{Z}=\mathrm{x}+2 \mathrm{y}\)

\(\mathrm{x}+2 \mathrm{y} \geq 100 \)

\(2 \mathrm{x}-\mathrm{y} \leq 0 \)

\(2 \mathrm{x}+\mathrm{y} \leq 200\)

\(\mathrm{x}, \mathrm{y} \geq 0\)

On solving equations \(2 \mathrm{x}-\mathrm{y}=0\) and \(\mathrm{x}+2 \mathrm{y}=100\) we get point \(\mathrm{B}(20,40)\)

On solving \(2 \mathrm{x}-\mathrm{y}=0\) and \(2 \mathrm{x}+\mathrm{y}=200\) we get \(\mathrm{C}(50,100)\)

\(\therefore\) Feasible region is shown by ABCDA

The corner points of the feasible region are:

\(\mathrm{A}(0,50), \mathrm{B}(20,40), \mathrm{C}(50,100), \mathrm{D}(0,200)\)

Let us evaluate the objective function \(\mathrm{Z}\) at each corner points:

At \(\mathrm{A}(0,50), \quad \mathrm{Z}=0+100=100\)

At \(\mathrm{B}(20,40), \mathrm{Z}=20+80=100\)

At \(\mathrm{C}(50,100), \mathrm{Z}=50+200=250\)

At \(\mathrm{D}(0, 200), \mathrm{Z}=0+400=400\)

So, Maximum value of \(\mathrm{Z}\) is 400.

It is given that \(Z=2 x+5 y\) and it is subject to constraints:

\(x+4 y \leq 24\)

\(3 x+y \leq 21\)

\(x+\) \(y \leq 9\)

\(x \geq 0\);

\(y \geq 0\)

Converting into equation format,

\(x+4 y = 24\) ...(1)

\(3 x+y = 21\) ...(2)

\(x+\) \(y = 9\) ...(3)

The corner points of this common region (shaded region) are: O(0,0), A(7,0), B(6,3), C(4,5), D(0,6).

x+4y = 24 and x+y = 9 intersects at C. From, (1) and (3) coordinates of C is (4,5).

3x+y = 21 and x+y = 9 intersects at B. From, (2) and (3) coordinates of B is (6,3).

Now, the following table shows that which corner point shows maximizing and minimizing value.

So, 33 is the maximum value of Z and it occurs at C.

Given:

\(Z=10 x+25y\)

Subject to: \(x \leq 3, y \leq 3, x+y \leq 5, x \geq 0, y \geq 0\)

First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.

Given:

\(\text { Max. } Z=-0.1 x_{1}+0.5 x_{2} \)

\(2 x_{1}+5 x_{2} \leq 80 \)

\(x_{1}+x_{2} \leq 20 \)

\(x_{1}, x_{2} \geq 0\)

Convert the inequality constraints into equations, we have

\(2 x_{1}+5 x_{2}=80 \) ...(1)

\(x_{1}+x_{2}=20\) ...(2)

\(2 \mathrm{x}_{1}+5 \mathrm{x}_{2}=80\) passes through the point \((0,16)\) and \((40,0)\).

\(x_{1}+x_{2}=20\) passes through the point \((0,20)\) and \((20,0)\).

Given:

\(Z=5 x+7 y\) and feaslble reglon \(O A B C\).

The corner points of the feasible region are \(O(0,0), A(7,0), B(3,4)\) and \(C(0,2)\). On evaluating the value of \(Z\), we get

\(\begin{array}{|l|l|} \hline \text{Corner points} & \text{Value of Z} \\ \hline O(0,0) & Z=5(0)+7(0)=0 \\ \hline A(7,0) & Z=5(7)+7(0)=35 \\ \hline B(3,4) & Z=5(3)+7(4)=43 \\ \hline C(0, 2) & Z=5(0)+7(2)=14 \\ \hline \end{array}\)

From the above table it's seen that the maximum value of \(Z\) is 43 .

Therefore, the maximum value of the function \(Z\) is 43 at \((3,4)\).

It is given that,\( z = x + y\) is subjected to condition:

\(x-y \leq-1\)

\(-x+y \leq 0\)

\(x, y \geq 0\)

Now, converting inequalities into an equation format.

\(x - y = -1\)...(i)

\( -x + y = 0\)...(ii)

The following table shows the values \(z\), when it is subjected to conditions (i) and (ii);

Given:

\(Z=3 x+2 y\) Subject to the constraints \(x+y \geq 8,3 x+5 y \leq 15\) and \(x \geq 0, y \leq 15\)

Converting the glven In equatlons Into the equatlons:

\(x+y=8 \ldots \)(1)

\(3 x+5 y=15 \)(2)

\(y=15 \ldots\)(3)

Reglon represented by \(x+y \geq 8\) :

The line \(x+y=8\) meets the coordinate axis at \(C(8,0)\) and \(D(0,8)\).

Table for \(x+y=8\)

\(\begin{array}{|l|l|l|} \hline X & 8 & 0 \\ \hline y & 0 & 8 \\ \hline \end{array}\)

\(\mathrm{A}(8,0) ; \mathrm{B}(0,8)\)

Join the points \(C\) and \(D\) to obtain the line.

We find that the point \((0,0)\) does not satisfy the in equation \(x+y>8\)

So, the region opposite to the origin represents the solution set to the in equation.

Reglon represented by \(3 x+5 y \leq 15:\)

The line \(3 x+5 y=15\) meets the coordinate axis at \(C(5,0)\) and \(D(0,3)\)

Table for \(3 x+5 y=15\)

\(\begin{array}{|l|l|l|} \hline X & 5 & 0 \\ \hline y & 0 & 3 \\ \hline \end{array}\)

\(C(5,0) ; \mathrm{D}(0,3)\)

Join the points \(C\) and \(D\) to obtain the line.

Clearly \((0,0)\) satisfies the in equation \(3 x+5 y \leq 15\).

So, the region containing the origin represents the solution set of this in equation.

Reglon represented by \(y \leq 15\) :

Line \(y=15\) is parallel to \(x\)-axis, its each point will satisfy the in equation in first quadrant.

So, its solution region will be towards origin.

Reglon represented by \(x \geq 0\) and \(y \geq 0\) :

Since every point in the first quadrant satisfies these in equations.

So the first quadrant is the region represented by the in equations \(x \geq 0\) and \(y \geq 0\).

There is no any common region for solution.

Given:

Maximize \(Z=3 x+5 y\) 

Subject to:

\(x+2 y \leq 20 \)

\(x+y \leq 15 \)

\(y \leq 5 \)

\(x, y \geq 0\)

We need to maximize \(\mathrm{Z}=3 x+5 y\)

First, we will convert the given inequations into equations, we obtain the following equations:

\(x+2 y=20, x+y=15, y=5, x=0\) and \(y=0\)

The line \(x+2 y=20\) meets the coordinate axis at \(A(20,0)\) and \(B(0,10)\). Join these points to obtain the line \(x+2 y=20\), \((0,0)\) satisfies the inequation \(x+2 y \leq 20\).

So, the region in \(x y\)-plane that contains the origin represents the solution set of the given equation.

The line \(x+y=15\) meets the coordinate axis at \(C(15,0)\) and \(D(0,15)\). Join these points to obtain the line \(x+y=15\). \((0,0)\) satisfies the inequation \(x+y \leq 15 .\) So, the region in \(x y\)-plane that contains the origin represents the solution set of the given equation.

\(y=5\) is the line passing through \((0,5)\) and parallel to the \(X\) axis.The region below the line \(y=5\) will satisfy the given inequation.

Region represented by \(x \geq 0\) and \(y \geq 0\) :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.

These lines are drawn using a suitable scale.

The corner points of the feasible region are \(O(0,0), C(15,0), E(10,5)\) and \(F(0,5)\) The values of \(Z\) at these corner points are as follows.

\(\begin{array}{|l|l|} \hline \text{Corner point} & Z=3 x+5 y \\ \hline O(0,0) & 3 \times 0+5 \times 0=0 \\ \hline C(15,0) & 3 \times 15+5 \times 0=45 \\ \hline E(10,5) & 3 \times 10+5 \times 5=55 \\ \hline F(0,5) & 3 \times 0+5 \times 5=25 \\ \hline \end{array}\)

We see that the maximum value of the objective function \(Z\) is 55 which is at \(E(10,5)\).

Thus, the optimal value of \(Z\) is 55.