Case study (4 Marks)

Question 14 Marks

A company has two factories located at P and Q and has three depots situated at A, B and C. The weekly requirement of the depots at A, B and C is respectively 5, 5 and 4 units, while the production capacity of the factories P and. Q are respectively 8 and 6 units. The cost (in ₹) of transportation per unit is given below.

Cost(in ₹)
TO/From	A	B	C
P	160	100	150
Q	100	120	100

Based on the above information, answer the following questions:
(i) Formulate the objective function and the constraints of the above Linear programming problem.
(ii) How many units should be transported from each factory to each depot in order that the transportation cost is minimum?

Answer

(i) Let the factory P supply x units per week to depot A and y units to depot B
so that it supplies 8−x−y units to depot C. Obviously $0 \leq x \leq 5,0 \leq y \leq 5,0 \leq 8-x-y \leq 4$.The given data can be represented diagrammatically as:

Thus, total transportation cost (in ₹)
$=160 x+100 y+150(8-x-y)+100(5-x)+120(5-y)+100(x+y-4)=10(x-7 y+190)$.
Hence the given problem can be formulated as an L.P.P as:
Minimize $Z=10(x-7 y+190)$
subject to the constraints
$\begin{array}{l}x+y \geq 4, \\ x+y \leq 8, \\ x \leq 5 \\ y \leq 5 \\ x \geq 0, y \geq 0\end{array}$
(ii) The feasible region corresponding to these in equations is shown shaded in the figure given below.

Corner Points	Value of $Z=10(x-7 y+190)$
A (4,0)	1940
B (5,0)	1950
C (5,3)	1740
D (3,5)	1580
E (0,5)	$1550 \rightarrow$ Minimum
F (0,3)	1690

We observe that Z is minimum at point E( 0, 5 ) and minimum value is ₹ 1550. Hence x = 0, y = 5 . Thus for minimum transportation cost, factory P should supply0, 5, 3 units to depots A, B, C respectively and factory Q should supply 5, 0, 1units respectively to depots A, B, C.

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Question 24 Marks

EQUATED MONTHLY INSTALMENTS (EMI): -
Each instalment can be considered as consisting of two parts:
(i) Interest on the outstanding loan
(ii) Repayment of part of the loan.
Methods of calculation of EMI or Instalment: -
EMI or Installment can be calculated by two methods:
1.Flat Rate Method
2. Reducing-balance method or Amortization of Loan
Rajesh purchased a house from a company for ₹2500000and made a down payment of ₹500000 He repays the balance in 25 years by monthly instalments at the rate of 9% per annum compounded monthly. $\left(\operatorname{Given}(1.0075)^{-300}=0.1062\right)$
Based on the above information, answer the following questions:
(i) Find the number of payments and find the rate of interest per month.
(ii) (a) What are the monthly payments of instalments using reducing balance method?
OR
(ii) (b) What are the monthly payments of instalments using flat rate method?
(iii) What is the total interest payment made in the process applied to calculate EMI in the above part 37 ( ii ) ?

Answer

(i) Here, time = 25 years
$\therefore$ Total number of payments $= 2 5 \times 1 2 = 3 0 0$
$R=9 \%$ per annum.
Rate of interest per month $=\frac{ 9 }{ 1 2 0 0 }= 0 . 0 0 7 5$
(ii) (a) Cost of house =₹ 2500000
Down Payment =₹500000
$\therefore$ Principal amount = ₹(2500000 - 500000)
=₹2000000
$
\begin{array}{l}
\text { EMI (using reducing balance method) }=\frac{P \times i}{1-(1+i)^{-n}} \\
=\frac{2000000 \times 0.0075}{1-(1+0.0075)^{-300}} \\
=\frac{15000}{1-(1.0075)^{-300}} \\
=\frac{15000}{1-(0.1062)} \\
=\frac{15000}{0.8938}=16782.27
\end{array}
$
Hence, monthly payment is ₹16782.27
OR
(ii) (b) Cost of house =₹2500000
Down Payment =₹500000
$\therefore$ Principal amount =₹(2500000 − 500000)
=₹2000000
$\begin{array}{l}\text { EMI (using flat rate method) }=P\left(i+\frac{1}{n}\right) \\ \qquad \begin{aligned} &
=2000000\left(0.0075+\frac{1}{300}\right)=2000000(0.0108333)\end{aligned}\end{array}$
= ₹21666.66
(iii) EMI (using reducing balance method) = ₹16782.27
$\therefore \quad$ Total int erest $=n \times$ EMI $-P$
$\begin{array}{l}=300 \times 16782.27-2000000 \\ =3034681\end{array}$
Hence, total interest is ₹3034681
When EMI is calculated by (using flat rate method), then
Total interest $=n \times E M I-P=300 \times 21666.6-2000000$
= ₹4499980

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Question 34 Marks

A student Shivam is running on a playground along the curve given by y = x²+7. Another student Manita standing at point ( 3, 7 ) on playground wants to hit Shivam by paper ball when Shivam is nearest to Manita.
(i) Let at any instant while running along the curve y = x²+7 Shivam’s position be ( x, y ) Find the expression for the distance ( D ) between Shivam and Manita in terms of ' x '.
(ii) Find the critical point(s) of the distance function.
(iii) (a) What is the distance between Shivam and Manita when they are at least distance from each other.
OR
(iii) (b) Find the position of Shivam, when he is closest to Manita.

Answer

(i) For all values of $x, y=x^2+7$
$\therefore$ Shivam's position at any point of $x$ will be $\left(x, x^2+7\right)$
The measure of the distance between Shivam and Manita, i.e., D
$
D=\sqrt{(x-3)^2+\left(x^2+7-7\right)^2}=\sqrt{(x-3)^2+x^4}
$
(ii) We have,
$
\begin{array}{l}
D=\sqrt{(x-3)^2+x^4} \\
\text { Let } \Delta=D^2=(x-3)^2+x^4
\end{array}
$
Now,
$
\begin{array}{l}
\frac{d}{d x}(\Delta)=2(x-3)+4 x^3=4 x^3+2 x-6 \\
\frac{d}{d x}(\Delta)=0 \Rightarrow x=1
\end{array}
$
(iii) (a): $\Delta^{\prime \prime}(x)=8 x^2+2$
Clearly, $\Delta^{\prime \prime}(x)=8 x^2+2>0$ at $x =1$
$\therefore$ Value of $x$ for which $D$ will be minimum is 1 .
For $x=1, y=8$.
Therefore, required distance $=D=\sqrt{(1-3)^2+(1)^4}=\sqrt{4+1}=\sqrt{5}$
OR
(iii) (b): $\Delta^{\prime \prime}(x)=8 x^2+2$
Clearly, $\Delta^{\prime \prime}(x)=8 x^2+2>0$ at $x=1$
$\therefore$ Value of $x$ for which $D$ will be minimum is 1 .
For $x =1, y = 8$.
Thus, the required position for Shivam is $( 1 , 8 )$ when he is closest to Manita.

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Applied Maths Case study (4 Marks)

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