Case study (4 Marks)

Question 14 Marks

Read the following text carefully and answer the questions that follow:
Linear programming is a method for finding the optimal values (maximum or minimum) of quantities subject to the constraints when a relationship is expressed as linear equations or inequations.
i. At which points is the optimal value of the objective function attained? (1)
ii. What does the graph of the inequality $3 x+4 y<12$ look like? (1)
iii. Where does the maximum of the objective function Z = 2x + 5y occur in relation to the feasible region shown in the figure for the given LPP? (2)

OR
What are the conditions on the positive values of p and q that ensure the maximum of the objective function Z = px + qy occurs at both the corner points (15, 15) and (0, 20) of the feasible region determined by the given system of linear constraints? (2? (2)

Answer

When we solve an L.P.P. graphically, the optimal (or optimum) value of the objective function is attained at corner points of the feasible region.
ii. From the graph of 3x + 4y < 12 it is clear that it contains the origin but not the points on the line 3x + 4y = 12

iii. Maximum of objective function occurs at corner points

Corner Points	Value of z = 2x + 5y
(0,0)	0
(7,0)	14
(6,3)	27
(4,5)	$33 \leftarrow$ Maximum
(0,6)	30

OR
Value of $Z=p x+q y$ at $(15,15)=15 p+15 q$ and that at $(0,20)=20 q$. According to given condition, we have $15 p +15 q =20 q \Rightarrow 15 p =5 q \Rightarrow q =3 p$

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Question 24 Marks

Read the following text carefully and answer the questions that follow:
Consider the following diagram, where the forces in the cable are given.

i. What is the equation of the line along cable $AD$ ? $(1)$
ii. What is length of cable $DC$ ? $(1)$
iii. Find vector $DB\ (2)$
OR
What is sum of vectors along the cable? $(2)$

Answer

$i.$ Clearly, the coordinates of $A$ are $(8, 10, 0)$ and $D$ are $(0, 0, 30)$
$\therefore$ Equation of $AD$ is given by
$\frac{x-0}{8-0}=\frac{y-0}{10-0}=\frac{z-30}{-30}$
$\Rightarrow \frac{z}{4}=\frac{y}{5}=\frac{30-z}{15}$
$ii$. The coordinates of point $C$ are $(15,20, 0)$ and $D$ are $(0, 0, 30)$
$\therefore$ Length of the cable $DC$
$=\sqrt{(0-15)^2+(0+20)^2+(30-0)^2}$
$=\sqrt{225+400+900}=\sqrt{1525}=5 \sqrt{61} m$
$iii.$ Since, the coordinates of point $B$ are $(-6, 4, 0) $ and $D$ are $(0, 0, 30),$
therefore vector $DB$ is
$(-6-0) \hat{i}+(4-0) \hat{j}+(0-30) \hat{k}$, i.e., $-6 \hat{i}+4 \hat{j}-30 \hat{k}$
OR
Required sum
$=(8 \hat{i}+10 \hat{j}-30 \hat{k})+(-6 \hat{i}+4 \hat{j}-30 \hat{k})+(15 \hat{i}-20 \hat{j}-30 \hat{k})$
$=17 \hat{i}-6 \hat{j}-90 \hat{k}$

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Question 34 Marks

Read the following text carefully and answer the questions that follow:
A shopkeeper sells three types of flower seeds $A_1 A_2 A_3$They are sold in the form of a mixture, where the proportions of these seeds are $4 : 4 : 2$ respectively. The germination rates of the three types of seeds are $45\%, 60\%$ and $35\%$ respectively.

Based on the above information:
$i.$ Calculate the probability that a randomly chosen seed will germinate. $(1)$
$ii.$ Calculate the probability that the seed is of type $A2,$ given that a randomly chosen seed germinates. $(1)$
$iii.\ A$ die is throw and a card is selected at random from a deck of $52$ playing cards. Then find the probability of getting an even number on the die and a spade card. $(2)$
$OR$
If $A$ and $B$ are any two events such that $P(A) + P(B) - P(A$ and $B) = P(A) ,$ then find $P( A |B). 2)$

Answer

$\text { Here, } P \left( E _1\right)=\frac{4}{10}, P \left( E _2\right)=\frac{4}{10}, P \left( E _3\right)=\frac{2}{10}$
$P\left(\frac{A}{E_1}\right)=\frac{45}{100}, P\left(\frac{A}{E_2}\right)=\frac{60}{100}, P\left(\frac{A}{E_3}\right)=\frac{35}{100}$
$\therefore P ( A )= P \left( E _1\right) \cdot P \left(\frac{A}{E_1}\right)+ P \left( E _2\right) \cdot P\left(\frac{A}{E_2}\right)+ P \left( E _3\right) \cdot P\left(\frac{A}{E_3}\right)$
$=\frac{4}{10} \times \frac{45}{100}+\frac{4}{10} \times \frac{60}{100}+\frac{2}{10} \times \frac{35}{100}$
$=\frac{180}{1000}+\frac{240}{1000}+\frac{70}{100}$
$=\frac{490}{1000}=4.9$
$\text ii.$ Required probability $ =P\left(\frac{E_2}{A}\right)$
$=\frac{P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)}{P(A)}$
$=\frac{\frac{4}{10} \times \frac{60}{100}}{\frac{490}{1000}}$
$=\frac{240}{490}=\frac{24}{49}$
$iii.$ Let,
$E_1=$ Event for getting an even number on die and
$E_2=$ Event that a spade card is selected
$\therefore P\left(E_1\right)=\frac{3}{6}$
$=\frac{1}{2}$
and $P\left(E_2\right)=\frac{13}{52}=\frac{1}{4}$
Then,$P \left( E _1 \cap E _2\right)= P \left( E _1\right) \cdot P \left( E _2\right)$
$=\frac{1}{2}, \frac{1}{4}=\frac{1}{8}$
$OR$
$P ( A )+ P ( B )- P ( A$ and $B )= P ( A )$
$\Rightarrow P(A)+P(B)-P(A \cap B)=P(A)$
$\Rightarrow P(B)-P(A \cap B)=0$
$\Rightarrow P(A \cap B)=P(B)$
$\therefore P ( A \mid B )=\frac{P(A \cap B)}{P(B)}$
$=\frac{P(B)}{P(B)}$
$=1$

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MATHS Case study (4 Marks)

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