Showing posts with label Maths. Show all posts
Showing posts with label Maths. Show all posts

Find two positive numbers whose sum is 16 and sum of whose cube is Minimum


Show that of all the rectangles inscribed in a circle of given radius . The Square has maximum Area.


Solutions


Let ABCD be rectangle which is  inscribed in a given circle of radius ‘r’
Show that of all the rectangles inscribed in a circle of given radius . The Square has maximum Area.
And Let θ be the angle between side of rectangle and Diameter of given circle.


Therefore from right angled  Î” ABC ,

We have 
  AB  = AC cosθ          ∵ AC = 2r
Let A(x) be the area of Rectangle ABCD
∴ A(x) = AB × BC
    A = (2r cos Î¸)(2r sin Î¸ )
    A =  4r2 sin Î¸ cos Î¸
    A = 2r2  (2sin Î¸ cos Î¸)
    A = 2r2  (sin 2θ )

⇒ 2r2 2 (cos 2θ ) = 0 ,As r2 is constant
⇒cos 2θ = 0
⇒cos 2θ =cos (Ï€/2)
⇒ Î¸ = Ï€/4
 =4r2  (-2sin 2θ 



∴ A has Maximum value at Î¸ = Ï€/4



Find two positive numbers whose sum is 16 and sum of whose cube is Minimum

Solution

Let us consider two numbers x and 16- x .
Then transforming our problem to mathematical form which says “sum of whose cube”  as follows
A (x) =   x3 + (16 - x)3…….. (1)
Differentiating both sides w .r. t  “x” , we get



     X = 8
So  x  =  8 will be the 1st required numbers if Double derivatives of A  w. r. t  ‘x’ comes to be positive at x = 8.
Differentiate (2)  w. r. t. ‘x’  .




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FINAL WORDS


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HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY MATRIX METHOD

How to find solutions of system of linear equations of three variables with  the help of Matrix Method. In this post we shall  solve such types of system of linear equations in which variables lies in the denominators.  Let us solve three linear equations of three variables given below by Matrix Method.


HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY  MATRIX  METHOD


These are not simple linear equations to solve. But these equations can be made simple by following substitution .



After substitution , these equations can be  reduced to following simple linear equations
2x + 3y + 4z = 3
3x - 4y + 5z = 5
  x + 2y - 3z = 6
To solve these equations by Matrix Method , Let us transform this set of linear equations to Matrix form

AX = B, and  =A-1 -----> (4)

where A is matrix comprises the coefficients  x , y and z respectively as shown in the matrix given below , And B is the matrix consist of constant terms on right hand side of these equations and X is the matrix of variables in the linear equations.
HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY  MATRIX  METHOD

1st of all we have to find A- and then product of the matrices A-1 and  as mentioned in equation (4). But to find inverse of matrix A, its determinant value must be non zero   ( Note it) .
Let us find determinant value of Matrix A as follows :-

|A| = 2[(-4)(-3)–(5)(2)]-3[(3)(-3)-(5)(1)]+4[(3)(2)-(-4)(1)] 

|A| = 2[12-10]-3[-9-5]+4[6+4] 
|A| = 2[2] -3[-14]+4[10] 
|A| = 4 +42+40 
|A| = 86 
Since the value of Determinant of Matrix a is non zero.
  ⇒    A-1  exists .
To find inverse ,we have to find ad joint of Matrix A . And to find Ad joint of any Matrix , we have to find Co factor Matrix . 

What is Co Factor Matrix


Co factor Matrix can be obtained by the co factors of all the elements of Matrix written in same place  where the element was written originally. 
Formula for Co factor of any element 
        Cij = (-1)i+jMij 


Co factor of A11 

Co factor of A11   element can be calculated by eliminating 1st row and 1st column and solving the remaining determinant.
(-1)1+1M11 
= 12 - 10 = 2


Co factor of A12 


Co factor of A12   element can be calculated   by eliminating 1st row and 2nd column and solving the remaining determinant.
(-1)1+2M12 


= -1(-9 - 5 )= 14

Co factor of A13 


Co factor of A13   element  can be  calculated   by eliminating 1st row and 3rd column and solving the remaining determinant.

(-1)1+3M13 
= 6 + 4 = 10

Co factor of A21 


Co factor of A21   element can be calculated   by eliminating 2nd row and 1st column and solving the remaining determinant.
(-1)2+1M11 

= -1(- 9 - 8) = 17

Co factor of A22 


Co factor of A22   element  can be calculated   by eliminating 2nd row and 2nd column and solving the remaining determinant.
(-1)2+2M22 
= - 6 - 4 = -10

Co factor of A23 


Co factor of A23   element can be calculated   by eliminating 2nd row and 3rd column and solving the remaining determinant.

(-1)2+3M23 
= -1(4 - 3) = -1

Co factor of A31


Co factor of A31   element can be calculated   by eliminating 3rd row and 1st column and solving the remaining determinant.

(-1)3+1M31 
 
= 15 + 16 = 31


Co factor of A32



Co factor of A32   element  can  be calculated   by eliminating 3rd row and 2nd column and solving the remaining determinant.

= (-1)3+2M32 
= -1(10 - 12) = 2


Co factor of A33



Co factor of A33   element  can  be calculated   by eliminating 3rd row and 3rd column and solving the remaining determinant.


(-1)3+3M33 
= - 8 - 9 = -17


Co Factor Matrix of A


Now write all the co factors calculated in Matrix Form as follows
HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY  MATRIX  METHOD

Ad joint Matrix of A


To find the Ad Joint of  Co factor Matrix, transform 1st row to 1st column, 2nd row to 2nd column  and 3rd rows to 3rd column as follows 


Formula to Find   A-1 



To Find the value of  A-1 ,Putting the value of Adj A in the formula given below.
HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY  MATRIX  METHOD


Find values of x ,y and z


Multiplying both the matrices which are on the right side.
HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY  MATRIX  METHOD
Simplifying the matrix so obtained
HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY  MATRIX  METHOD
Dividing each element by 86 to get matrix of 3×1 (i. e. Perform scalar multiplication of matrix ).
HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY  MATRIX  METHOD

Using the equality of two matrices .We can write the values of x, y  and z respectively like this

 x =277/86  , y = 4/86  and z = -77/86

Now the values of P, Q and R can be calculated by putting the values of x , y and z in equations (1) , (2 ) and (3) respectively.

P = 86/277  ,Q = 86/4   and R = -86/77

Don't Forget to Watch this Video of same Method

Verification of solution



Putting values of P , Q and R in given  equations , these values must satisfies  three given equations
From 1st equation

HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY  MATRIX  METHOD

Similarly other two equations can be verified

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Conclusion


I have discussed the method of solving the System of linear Equations with the help of Matrix Method , method to find inverse of a matrix ,How to find inverse of 3×3 Matrix,how to solve determinant,how to solve system of equation by matrix, matrix method of solving system of equations of three variables,If you liked the post Don't  forget to share it with your friends , And in case of any improvement please make use of Comment Box . 

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How should the wire of 28 m be cut so that the combined area of the circle and square is as small as possible ?

Application of Derivative 

A piece of wire 28 cm long is to be cut into two pieces. One piece is to be made into a circle and another into a square. How should the wire be cut so that the combined area of the two figures is as small as possible?

Let the wire be cut at a distance of  x meter  from one end. Therefore then two pieces of wire be x m and (28-x) m.


Calculate Dimension of Circle and Square


Now 1st part be turned into a square and  the 2nd part be be made into a circle.

Since 1st part of the wire is turned into square. then its perimeter will be x m. 
So using formula of perimeter of square , we can calculate side of the square = x/4 m


Calculate Areas of Circle and Square


Therefore Area of square = (x/4)(x/4) sq m

                                     A1 = x2/16


And  when 2nd part of the wire is turned to circle, then its perimeter ( circumference ) will be 28 - x m. So using formula of perimeter of square , And if  "r" be  radius of the circle , Then
Circumference of circle =  2 Ï€ r =  (28-x)
 ∴  r = (28-x)/2Ï€

We know that Area of Circle A2   = Ï€ r2  

                                     A2  Ï€[(28-x)/2Ï€]2  


Express Areas in terms of Function





To find value/s of x


Now to find the value of x for which this function A(x) is maximum or minimum ,put A(x) = 0



To Test the Minimum Value of  Function


Now we have the value of "x" on which either A(x) have maximum or minimum value . To check the maximum or minimum value we have to find A''(x) as follows






So A''(x) has positive value Therefore A(x) shall have maximum value at x = 112/(Ï€ + 4)

Hence two pieces of wire should be of length x m and (28-x) m

These pieces should be of length 112/(Ï€+4) and 28Ï€/(Ï€ + 4)


Verification



we can calculate the sum of these pieces , it must be 28 m


1st part     

   
112/(Ï€+4) = 112/{(22/7)+4}=112×7/50 = 784/50


2nd part 


28Ï€/(Ï€ + 4) = {28×22/7}/{(22/7)+4} = 88×7/50 = 616/50

Sum of Two Parts 


 112×7/50 + 28×7/50 = (784+616)/50
                                                                 
  = 1400/50= 28 m



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Conclusion



Thanks for visiting this website and spending your valuable time to read this post regarding How should the wire of 28 m be cut so that the combined area of the circle and square is as small as possible , s .If you liked this post , don't forget to   share it with your friends to benefit them also ,we shall meet in next post , till then bye and take care......



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HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY MATRIX METHOD

Discussed  short cut for inverse of matrix and solving linear equations of three variables .
Consider three linear equations of three variables .

2x + 3y - 4z = 10

3x - 2y + 4z = 12

  x  -  y  +  z  =  5
Changing these equations into Matrix Form like this

AX = B
X =  A-1 B - - - - -(1)

where A is Matrix of 3×3 order , which consist of  Coefficients of  x ,y  and z respectively. X is the matrix of order 3×1 and whose elements are  variables  in given linear equations . B is the matrix of 3×1 and consist of  the constant terms on the right hand sides of all the equations such that


How to Find A-1 

To find A-1 ,we have to check the determinant value of Matrix A, if the Determinant value of Matrix A is non Zero then  A-1 Exists, otherwise not.

How to find Determinant  Det |A|

|A| = 2(2) + 3(1) - 4(-1)

|A| = 2(2) + 3(1) - 4(-1)

|A| = 4 + 3 +4

|A| =  7
Since |A| is non Zero therefore A-1 Exists

To find Co factors of elements  and Ad joint Matrix of A

1st of  all  put all the elements of matrix A in 3 rows and 3 columns as written in matrix A then copy the 1st and 2nd columns as  4th and 5th columns , After this we have 3×5 arrangement as shown is 1st  figure given below, Now  complete the arrangement as 5×5 by copying 1st row and 2nd row as 4th and 5th row respectively. It can be seen in 2nd figure given below.

Find Co factors of A11  element and write it in C11   position


The element whose co factor is to be find out , is marked in red. and the co factor will be calculated by eliminated that row and column in which red coloured element is lying, Here co factor will be calculated by cross multiplication of  purple coloured four elements

Find Co factors of  A12  element and write it in C12   position

Here co factor will be calculated by cross multiplication of  purple coloured four elements

Find Co factors of  A13  element and write it in C13   position

Here co factor will be calculated by cross multiplication of  purple coloured four elements

Find Co factors of A21  element and write it in C21   position




Find Co factors of A22  element and write it in C22   position




Find Co factors of  A23  element and write it in C23   position




Find Co factors of A31  element and write it in C31   position


Find Co factors of A31  element and write it in C31   position



Find Co factors of A31  element and write it in C31   position


How to find values of  x , y and z


Therefore ad joint of Matrix A can be written as below


Now using the property of equality of two matrices ,
x   =   52/22   
y   =   -18/11 and 
z    =  -15/11


Verification 

we can check whether the values of x , y and z so calculated satisfies our system of linear equation by putting their values in one or all the given equations.

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Do not Forget to watch this video of same Problem

You can  clear your doubts if any after watching this video





Conclusion

Thanks for visiting this website and spending your valuable time to read this post regarding HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY MATRIX METHOD , short cut for inverse of matrix .If you liked this post , don't forget to   share it with your friends to benefit them also ,we shall meet in next post , till then bye and take care......



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