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NAME ID NO.

PRODIP SOMODDER 15-087

MD. SABUJ MIAH 15-107

MD. MOINUL ALAM 15-137

MD. KAMRUL HASAN SHOHEL 15-171

SANJOY KUMAR HALDAR 15-251

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` Matrix and basic function of Matrix,` Solving of business related matrix problem,` Differentiation,` Maximum and minimum of the function,` Solving of business related problem &` Conclusion.

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` A matrix can be a square or rectangular array of values,enclosed in brackets. In other words, a matrix is a rectangular array of mathematical elements, e.g. the coefficients of linear equations, whose rows and columns can be combined with thoseof other arrays to solve problems. Example of matrix includes:

3 5 7 A= 4 6 2

9 3 4 33

33

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` Square Matrix` Row and Column Matrix` Diagonal Matrix` Unit Matrix` Zero Matrix &` Scalar Matrix

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T here are three basic functions of matrix. T hese are-

1. Matrix addition2. Matrix subtraction3. Matrix multiplication

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` Cramers rule` Inverse of a matrix` Determinant` Minor matrix` Cofactor matrix` Adjoint matrix

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Solution:

Let, we obtained a set of linear equation,

3x + 5y - 4 z = 60002x - 3y + z = 5000-x + 4y + 6y = 13000

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W e can arrange the given information under matrix form as follows-

=

W e can find the value of A 1 as follows:

F irstly, we have to f ind the d e te rmin a nt o f A .

lAl = 3(-18 - 4) - 5(12 + 1) - 4( 8 - 3)= -66 - 65 - 20= -151

S ec ondly, we have to f ind the minor m a trix

M=

641

132

453

32

53

12

43

13

45

41

53

61

43

64

4541

32

61

12

64

13

13000

5000

6000

Z

Y

X

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T h irdly, we have to f ind the c o- fac tor

c=

F ourt h ly, we have to f ind the A djoint o f A

Adj A =

F in a lly, we have to f ind the in ve rs e o f A

A1 = Adj A

=1/-151

19117

171446

51322

19175

111413

74622

19175

111413

74622

A

1

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`

=

=

=

Ans we r: bread( X) = 3000 , biscuit (Y) = 1000 , cake (Z) = 2000.

Z

Y

X

15119

15117

1515 151

11

151

14

151

13151

715146

15122

15113000)13(5000)17(6000)5(

151 13000)11(5000)14(6000)13(

15113000)7(5000)46(6000)22(

20001000

3000

13000

5000

6000

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` Differentiation or the derivative is theinstantaneous rate of change of a function withrespect to one of its variables. T his is equivalent to

finding the slope of the tangent line to the functionat a point.

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1. Simpl e P o we r Rul eIf we have a function of f ( x ) = x n where n is a integer.

dx ndx = n x n-1

2. P rodu c t Rul e

T he chain rule allows us to differentiate, a product of termsthat depend on the differentiation variable. For example,U( x) and V ( x) are functions that depend on x . T hen thedifferential is given by:

d dx (U V ) = U dVdx + V d U dx

3. Q uoti e nt Rul eW here we have a quotient, the rule for differentiation isd dx (U/V) = [U dVdx - V dUdx ]/ V 2

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` A local maximum (or minimum) at a point is the largest(or smallest) value of the function in the vicinity of thatpoint; an absolute maximum (or minimum) is thelargest (or smallest) value of the function over all

values for which the function is defined.` One of the most useful applications of calculus in

management and economics is finding the maximumor minimum values for a function. To find out theseoptimum values, we need to develop a keener senseof the behavior of the graph of a function.

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` In calculus, the first derivative test determines whether a givencritical point of a function is a maximum, a minimum, or neither.

` Critical points are those points that are candidates for maximum or minimum values.

` There are three kinds of Critical points: stationary points,cusps, and endpoints .

C riti ca l P oints

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` T he slope of a line tangent to a curve at a point is,by definition, the value of the first derivative at thatpoint .hence when the first derivative is 0, the

tangent line is horizontal. Points where this occursare called stationary points.

` Def inition: A stationary point on f(x) is appoint

where f(x)=0

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Solution: A(x) =100/x+0.04x+1Or, A(x) =100x -1+0.04x+1

A(x) =-100x -2+0.04 At stationary points, A(x)=0

-100x -2+0.04=0x2 =100/0.04

x=50,-50

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x=-50, it is not possible because gallons cannot be negative.The coordinates of the stationary pointA(50) =100/50+0 . 04 ( 50)+1

=5Coordinates ( 50,5)For calculating the point; ma ximum or minimumA(+49) =-100 ( 49) -2 +0 . 04

=-0 . 001 ; f( x) is decreasing

A(+51) =-100 ( 51) -2 +0 . 04=+0 . 002 ; f( x) is increasing

So , this point is a local minimum.

The minimum average cost per gallonA(50) =100/50+0 .04 ( 50)+1 =5$ per gallons

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` In calculus, a branch of mathematics, the s ec ond d e riva tive te st is a criterion often useful for determining whether a given stationary point of a

function is a local maximum or a local minimum.` T he second derivative test may also be used to

determine the concavity of a function as well as afunction's points of inflection.

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Solution

P(x)=(100+10x) 1/2 -0.2x

P (x)=1/2(100+10x) -1/2 .10-0.2=5(100+10x) -1/2 -0.2

P (x)=5(-1/2)(100+10x) -3/2 .10=-25(100+10x) -3/2

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At a stationary point , P (x)=0,

5(100+10x) -1/2 -0.2=0(100+10x) 1/2 =5/0.2100+10x=25 2

10x=625-100x=52.5

p(52.5)=-25/(100+10X52.5) 3/2

=-0.0016

p(52.5)

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` Matrix and differential calculus are very importanttools of business math.

` T his two tools are used to find price, output, and

profit and analyze input & output.

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