Exercise 12.1 (Revised) - Chapter 12 - Linear Programming - Ncert Solutions class 12 - Maths

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Chapter 12: Linear Programming - NCERT Solutions Class 12 Maths

Solve the following Linear Programming Problems graphically:
Ex 12.1 Question 1.

Maximize $\mathbf{Z}=3 x+4 y$ subject to the constraints: $x+y \leq 4, x \geq 0, y \geq 0$.

Answer.

As $x \geq 0, y \geq 0$, therefore we shall shade the other inequalities in the first quadrant only.

Now $x+y \leq 4$

Let $x+y=4$
$
\Rightarrow \frac{x}{4}+\frac{y}{4}=1
$

Thus the line has 4 and 4 as intercepts along the axes. Now, $(0,0)$ satisfies the inequation, i.e., $0+0 \leq 4$. Therefore, shaded region $O A B$ is the feasible solution.

Its corners are $\mathrm{O}(0,0), \mathrm{A}(4,0), \mathrm{B}(0,4)$
At $O(0,0) \quad Z=0$
At $\mathrm{A}(4,0) \quad \mathrm{Z}=3 \times 4=12$
At $B(0,4) \quad Z=4 \times 4=16$

Hence, $\max \mathrm{Z}=16$ at $x=0, y=4$.
Ex 12.1 Question 2.

Minimize $\mathbf{Z}=-3 x+4 y$ subject to $x+2 y \leq 8,3 x+2 y \leq 12, x \geq 0, y \geq 0$.

Answer.

Consider $x+2 y \leq 8$
Let $x+2 y=8$
$
\begin{aligned}
& \Rightarrow \frac{x}{8}+\frac{y}{4}=1 \\
& \therefore a=8, b=4
\end{aligned}
$

Since, $(0,0)$ satisfies the inequaitons $x+2 y \leq 8$
Therefore, its solution contains $(0,0)$
Again $3 x+2 y \leq 12$
Let $3 x+2 y=12$
$
\Rightarrow \frac{x}{4}+\frac{y}{6}=1
$

Again, $(0,0)$ satisfies $3 x+2 y \leq 12$

Therefore its solution contains $(0,0)$.
The feasible region is the solution set which is double shaded and is OABCO.
At $O(0,0) \mathrm{Z}=0$
At $\mathrm{A}(4,0) \mathrm{Z}=-3 \times 4=-12$
At $B(2,3) Z=-3 \times 2+4 \times 3=6$
At $C(0,4) Z=4 \times 4=16$
Hence, minimum $\mathrm{Z}=-12$ at $x=4, y=0$.

Ex 12.1 Question 3.

Maximize $\mathbf{Z}=5 x+3 y$ subject to $3 x+5 y \leq 15,5 x+2 y \leq 10, x \geq 0, y \geq 0$.

Answer.

We first draw the graph of equation $3 x+5 y=15$
$
\Rightarrow x=\frac{15-5 y}{3}
$

For $y=3, x=0$
And for $y=0, x=5$

$\text { Similarly, for equation } 5 x+2 y=10 \text {, the points are }(2,0) \text { and }(0,5) \text {. }$

As $(0,0)$ satisfies both the inequations and also $x \geq 0, y \geq 0$, then the feasible require contains the half-plane containing $(0,0)$.

Therefore, the feasible portion is OABC which is shown as shaded in the graph.

Co-ordinates of point $\mathrm{B}$ can be obtained by solving $3 x+5 y=15$ and $5 x+2 y=10$ and it is
$
\text { B }\left(\frac{20}{19}, \frac{45}{19}\right) \text {. }
$

Thus, co-ordinates of $\mathrm{O}, \mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are $(0,0),(2,0),\left(\frac{20}{19}, \frac{45}{19}\right)$ and $(0,3)$.
$
\begin{aligned}
& \mathrm{Z}=5 x+3 y=0 \quad \text { (if } x=0, y=0) \\
& \mathrm{Z}=5 \times 2+3 \times 0=10 \quad(\text { if } x=2, y=0) \\
& \mathrm{Z}=5 \times \frac{20}{19}+3 \times \frac{45}{19}=\frac{235}{19} \quad\left(\text { if } x=\frac{20}{19}, y=\frac{45}{19}\right. \text { ) } \\
& \mathrm{Z}=5 \times 0+3 \times 3=9 \quad(\text { if } x=0, y=3)
\end{aligned}
$

Hence, $\mathrm{Z}=\frac{235}{19}$ is maximum when $x=\frac{20}{19}, y=\frac{45}{19}$.
Ex 12.1 Question 4.

Minimize $\mathbf{P}=3 x+5 y$ such that $x+3 y \geq 3, x+y \geq 2, x, y \geq 0$.

Answer.

For plotting the graphs of $x+3 y=3$ and $x+y=2$, we have the following tables:

The feasible portion represented by the inequalities $x+3 y \geq 3, x+y \geq 2$ and $x, y \geq 0$ is $\mathrm{ABC}$ which is shaded

in the figure. The coordinates of point $\mathrm{B}$ are $\left(\frac{3}{2}, \frac{1}{2}\right)$
Which can be obtained by solving $x+3 y=3$ and $x+y=2$.
At A $(0,2)$
$
\mathrm{Z}=3 \times 0+5 \times 2=10
$

At $\mathrm{B}\left(\frac{3}{2} ; \frac{1}{2}\right)$
$
\mathrm{Z}=3 \times \frac{3}{2}+5 \times \frac{1}{2}=\frac{9}{2}+\frac{5}{2}=\frac{14}{2}=7
$

At $\mathrm{C}(3,0)$
$
\mathrm{Z}=3 \times 3+5 \times 0=9
$

Hence, $\mathrm{Z}$ is minimum is 7 when $x=\frac{3}{2}$ and $y=\frac{1}{2}$.
Ex 12.1 Question 5.

Maximize $\mathbf{Z}=3 x+2 y$ subject to $x+2 y \leq 10,3 x+y \leq 15, x, y \geq 0$.

Answer.

Consider $x+2 y \leq 10$
Let $x+2 y=10$
$
\Rightarrow \frac{x}{10}+\frac{y}{5}=1
$

Since, $(0,0)$ satisfies the inequation, therefore the half plane containing $(0,0)$ is the required plane.

Again $3 x+2 y \leq 15$
Let $3 x+y=15$
$\Rightarrow \frac{x}{5}+\frac{y}{15}=1$
It also satisfies by $(0,0)$ and its required half plane contains $(0,0)$.
Now double shaded region in the first quadrant contains the solution.
Now OABC represents the feasible region.
$
\mathrm{Z}=3 x+2 y
$

At $O(0,0) \quad Z=3 \times 0+2 \times 0=0$
At $A(5,0) \quad Z=3 \times 5+2 \times 0=15$
At $B(4,3) \quad Z=3 \times 4+2 \times 3=18$
At $C(0,5) \quad Z=3 \times 0+2 \times 5=10$

Hence, $\mathrm{Z}$ is maximum i.e., 18 at $x=4, y=3$.
Ex 12.1 Question 6.

Minimize $\mathbf{Z}=x+2 y$ subject to $2 x+y \geq 3, x+2 y \geq 6, x, y \geq 0$. Show that the minimum of $Z$ occurs at more than two points.

Answer.

Consider $2 x+y \geq 3$
Let $\quad 2 x+y=3 \Rightarrow y=3-2 x$

$(0,0)$ is not contained in the required half plane as $(0,0)$ does not satisfy the inequation $2 x+y \geq 3$.

Again $x+2 y \geq 6$
Let $x+2 y=6$

$
\Rightarrow \frac{x}{6}+\frac{y}{3}=1
$

Here also $(0,0)$ does not contain the required half plane. The double shaded region XABY is the solution set. Its corners are A $(6,0)$ and $\mathrm{B}(0,3)$.

At $A(6,0) Z=6+0=6$
At $B(0,3) Z=0+2 \times 3=6$
Therefore, at both points the value of $Z=6$ which is minimum. In fact at every point on the line $\mathrm{AB}$ makes $\mathrm{Z}=6$ which is also minimum.
Ex 12.1 Question 7.

Minimize and Maximize $\mathrm{Z}=5 x+10 y$ subject to
$
x+2 y \leq 120, x+y \geq 60, x-2 y \geq 0, x, y \geq 0 \text {. }
$

Answer.

Consider $x+2 y \leq 120$
Let $x+2 y=120$
$
\Rightarrow \frac{x}{120}+\frac{y}{60}=1
$

The half plane containing $(0,0)$ is the required half plane as $(0,0)$ makes $x+2 y \leq 120$, true.

Again $x+y \geq 60$
Let $x+y=60$
Also the half plane containing $(0,0)$ does not make $x+y \geq 6$ true.
Therefore, the required half plane does not contain $(0,0)$.
Again $x-2 y \geq 0$
Let $x-2 y=0 \Rightarrow x=2 y$
Let test point be $(30,0)$.

$\Rightarrow x-2 y \geq 0 \Rightarrow 30-2 \times 0 \geq 0$ It is true.
Therefore, the half plane contains $(30,0)$.
The region CFEKC represents the feasible region.
At $\mathrm{C}(60,0) \quad \mathrm{Z}=5 \times 60=300$
At $\mathrm{F}(120,0) \quad \mathrm{Z}=5 \times 120=600$
At $\mathrm{E}(60,30) \quad \mathrm{Z}=5 \times 60+10 \times 30=600$

At $\mathrm{K}(40,20) \quad \mathrm{Z}=5 \times 40+10 \times 20=400$
Hence, minimum $\mathrm{Z}=300$ at $x=60, y=0$ and maximum $\mathrm{Z}=600$ at $x=120, y=0$ or $x=60, y=30$.
Ex 12.1 Question 8.

Minimize and Maximize $\mathrm{Z}=x+2 y$ subject to $x+2 y \geq 100,2 x-y \leq 0,2 x+y \leq 200, x, y \geq 0$.

Answer.

Consider $x+2 y \geq 100$

Let $x+2 y=100 \Rightarrow \frac{x}{100}+\frac{y}{50}=1$ $x+2 y \geq 100$ represents which does not include $(0,0)$ as it does not made it true.

Again consider $2 x-y \leq 0$
Let $2 x-y=0 \Rightarrow y=2 x$
Let the test point be $(10,0)$.
$\therefore 2 \times 10-0 \leq 0$ which is false.

Therefore, the required half does not contain $(10,0)$.
Again consider $2 x+y \leq 200$
Let $2 x+y=200$
$
\Rightarrow \frac{x}{100}+\frac{y}{200}=1
$

Now $(0,0)$ satisfies $2 x+y \leq 200$

Therefore, the required half place contains $(0,0)$.
Now triple shaded region is ABCDA which is the required feasible region.
At $\mathrm{A}(0,50)$
$
\mathrm{Z}=x+2 y=0+2 \times 50=100
$

At $B(20,40) Z=20+2 \times 40=100$
At $C(50,100) Z=50+2 \times 100=250$
At $D(0,200) Z=0+2 \times 200=400$
Hence maximum $\mathrm{Z}=400$ at $x=0, y=200$ and minimum $\mathrm{Z}=100$ at $x=0, y=50$ or $x=20, y=40$.
Ex 12.1 Question 9.

Maximize $\mathbf{Z}=-x+2 y$ subject to the constraints: $x \geq 3, x+y \geq 5, x+2 y \geq 6, y \geq 0$.

Answer.

Consider $x \geq 3$
Let $x=3$ which is a line parallel to $y$-axis at a positive distance of 3 from it.

Since $x \geq 3$, therefore the required half-plane does not contain $(0,0)$.
Now consider $x+y \geq 5$
Let $x+y=5$
$
\Rightarrow \frac{x}{5}+\frac{y}{5}=1
$

Now $(0,0)$ does not satisfy $x+y \geq 5$, therefore the required half plane does not contain $(0$, $0)$.

Again consider $x+2 y \geq 6$
Let $x+2 y=6$
$
\Rightarrow \frac{x}{6}+\frac{y}{3}=1
$

Here also $(0,0)$ does not satisfy $x+2 y \geq 6$, therefore the required half plane does not contain $(0,0)$.

The corners of the feasible region are A $(6,0), \mathrm{B}(4,1)$ and $\mathrm{C}(3,2)$.
At $A(6,0) \mathrm{Z}=-6+2 \times 0=-6$

At $B(4,1) \mathrm{Z}=-4+2 \times 1=-2$
At $C(3,2) Z=-3+2 \times 2=1$
Hence, maximum $\mathrm{Z}=1$ at $x=3, y=2$.
Ex 12.1 Question 10.

Maximize $\mathbf{Z}=x+y$ subject to $x-y \leq-1,-x+y \leq 0, x, y \geq 0$.

Answer.

Consider $x-y \leq-1$
Let $x-y=-1$
$\Rightarrow x=y-1$

If $(0,0)$ is the test point then $x-y \leq-1 \Rightarrow 0 \leq-1$ which is false and thus the required plane does not include $(0,0)$.

Again $-x+y \leq 0$
Let $-x+y=0$
$
\Rightarrow y=x
$

$\text { For }(1,0)-1 \leq 0 \text { which is true, therefore the required half-plane include }(1,0) \text {. }$

It is clear that the two required half planes do not intersect at all, i.e., they do not have a common region.

Hence there is no maximum $\mathrm{Z}$.