Maths

Minimize and Maximise $z=5x+10y$

Subject to $x+2y\le 120,x+y\ge 60,x-2y\ge 0,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+2y=120\\ x+y=60\\ x-2y=0\\ x,y=0\end{array}$

$\Rightarrow \frac{x}{120}+\frac{y}{60}=120$

$\begin{array}{l}\frac{x}{60}+\frac{y}{60}=1\\ x=2y\\ x,y=0\end{array}$

The graph of the inequalities in shown.

The shaded founded region ABCD is the feasible region with the corner points $A\left(60,0\right),B\left(120,0\right),C\left(60,30\right)\&D\left(40,20\right)$

The value of Z a these corner points are

The minimum value of Z is 300 at (60,0) and the maximum value of Z is 600 at all the points on the line segment joining B (120,0) and C (60,30).

Minimize $z=x+2y$

Subject to $2x+y\ge 3,x+2y\ge 6,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}2x+y=3\\ x+2y=6\\ x,y\ge 0\end{array}$

$\Rightarrow \frac{x}{\frac{3}{2}}+\frac{y}{3}=1$

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

$x,y\ge 0$

The feasible region is unbounded the corner point are A (6,0), B (0,3)

The value of Z at these corner points are follows.

Since, the feasible region is unbounded, a graph of $x+2y<6$ is drawn.

Also since there is no point common in feasible region and region $x+2y<6$ .

$z=6$ is maximum on all points joining line (0,3), (6,0)

i.e,  $z=6$ will be minimum on $x+2y=6$ 

Maximum $z=3x+2y$

Subject to $x+2y\le 10,3x+y=\le 15,x,y\ge 0$

The corresponding equation of the given inequalities are :

$\begin{array}{l}x+2y=10\\ 3x+y=15\\ x,y=0\end{array}$

$\Rightarrow \frac{x}{10}+\frac{y}{5}=1$

$\begin{array}{l}\frac{x}{5}+\frac{y}{5}=1\\ x,y=0\end{array}$

The graph of the given inequalities

The shaded bounded region OABC in the feasible region with the corner points

$O(0,0),A(5,0),B(4,3),C(0,5)$

The value of Z at these points are given below.

Therefore, the maximum value of Z is 18 at point (4,3).

Minimize $z=3x+5y$

Such that $x+3y\ge 3,x+y\ge 2,x,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+3y=3\\ x+y=2\\ x,y=0\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{3}+\frac{y}{1}=1\\ \frac{x}{2}+\frac{y}{2}=1\\ x,y=0\end{array}$

The graph of the given inequalities is

The feasible region is unbounded. The corner points are $A\left(3,0\right),B\left(\frac{3}{2},\frac{1}{2}\right)\&C\left(0,2\right)$

The values of Z at these corner points as follows.

As the feasible region is unbounded, 7 may or may not be minimum value of Z.

We draw the graph of inequality $3x+5y<7$ .

The feasible region has no common point with $3x+5y<7$ .

Therefore minimum value of Z in 7 at $B\left(\frac{3}{2},\frac{1}{2}\right)$

Maximise $z=5x+3y$

Subject to $2x+5y\le 15,5x+2y\le 10,x\ge 0,y\ge 0$

The corresponding equation of the above linear inequalities are

$\begin{array}{l}3x+5y=15\\ 5x+2y=10\&x=0,y=0\end{array}$

$\Rightarrow \frac{x}{5}+\frac{y}{3}=1$

$\begin{array}{l}\frac{x}{2}+\frac{y}{5}=1\\ x=0,y=0\end{array}$

The graph of its given inequalities.

The shaded region OABC is the feasible region which is bounded with the corner points

$O\left(0,0\right),A\left(2,0\right),B\left(\frac{20}{19},\frac{45}{19}\right)\&C\left(0,3\right)$

The values of Z at these points are

Therefore the maximum value of Z is $\frac{235}{19}at,B\left(\frac{20}{19},\frac{45}{19}\right)$

Minimize $z=-3x+4y$

Subject to $x+2y\le 8,3x+2y\le 12,x\ge 0,y\ge 0$

The corresponding equation of the given inequalities are

$\begin{array}{l}x+2y=8\\ 3x+2y=12\\ x=0,y=0\end{array}$

$\Rightarrow \frac{x}{8}+\frac{y}{4}=1$

$\begin{array}{l}\frac{x}{4}+\frac{y}{6}=1\\ x=0,y=0\end{array}$

The graph is shown below.

The bounded region OABC is the feasible region with the corner points O (0,0), A (4,0), B (2,3), and C (0,4

The value of Z at these points are

Therefore, the minimum value of Z is -12 at (4,0).

Maximise $z=3x+4y$

Subject to the constraints: $x+y\le 4,x\ge 0,y\ge 0$

The corresponding equation of the above inequality are

$\begin{array}{l}x+y=4\\ x=0,y=0\end{array}$

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

$x=0,y=0$

The graph of the given inequalities.

The shaded region OAB is the feasible region which is bounded.

The corresponding of the corner point of the feasible region are O (0,0), A (4,0), and B (0,4).

The value of Z at these points are as follows,

Corner point $z=3x+4y$

O (0,0) 0

A (4,0) 12

B (0,4) 16

Therefore the maximum value of Z is 16 at the point B (0,4).

$\hat{i}\left(\hat{j}*\hat{k}\right)+\hat{j}\left(\hat{i}*\hat{k}\right)+\hat{k}\left(\hat{i}*\hat{j}\right)$

$\begin{array}{l}=\hat{i}.\hat{i}+\hat{j}\left(-\hat{j}\right)+\hat{k}.\hat{k}\\ =1-\hat{j}.\hat{j}+1\\ =1-1+1\\ =1\end{array}$

Therefore, the correct answer is (C)

Let, $\overrightarrow{a}\&\overrightarrow{b}$ be two unit vectors and $\theta$ be the angle between them.

Then, $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=1$

Now, $\overrightarrow{a}+\overrightarrow{b}$ is a unit vector if $\left|\overrightarrow{a}+\overrightarrow{b}\right|=1$

$\begin{array}{l}\left|\overrightarrow{a}+\overrightarrow{b}\right|=1\\ {(\overrightarrow{a}+\overrightarrow{b})}^2=1\\ (\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}+\overrightarrow{b})=1\\ \overrightarrow{a}.\overrightarrow{a}+\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{a}+\overrightarrow{b}.\overrightarrow{b}=1\\ {\left|\overrightarrow{a}\right|}^2+2\overrightarrow{a}.\overrightarrow{b}+{\left|\overrightarrow{b}\right|}^2=1\\ 1^2+2\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|cos\theta +1^2=1\end{array}$

$1+2.1.1cos\theta +1=1$ [ $\therefore$ $\overrightarrow{a}\&\overrightarrow{b}$ is unit vector.]

$\begin{array}{l}2cos\theta =1-2\\ cos\theta =\frac{-1}{2}=\frac{2\pi }{3}\\ \therefore \theta =\frac{2\pi }{3}\end{array}$

Therefore, the correct answer is (D)