Group-J 10
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Transcript of Group-J 10
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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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