KACL-DS-INC

A recrangular board is given to form a cuboid for a

--> container (storage use). The maximum area needed is 298 m^2

--> (square meter). The dimension must be (3X + 5) m by

--> (3X + 4) m. However, a distance of X/2 must be cut (extracted)

--> from each edge before this shape could be form.

--> 1] Determine the equation that will represent this shape in

--> a parabolic form.

--> 2] Find the value of X that will be extracted(removed) from

--> each edge. What is the length of the actual board.

--> 3] What is the value of the extrema .

--> 4] Find F'(X) and F(X) intersection point i.e what is the

--> intersection point of DY/DX and F(x).

--> 5] What will be the value of this shape after full formation.

--> 6] What will the area be if X is to be 3.5 in value.

--> 7] Find the range of the function derived and

--> graph F(X) and Dy/Dx.

--> Produced by MATCAL Program

--> (I)

--> D1 = ( 3X + 5 ) , Dimension of one side of the board

--> D2 = ( 3X + 4 ) , Dimension of the other side of the board

--> A cut of X/2 is to be removed from each side of the board

--> D1 = ( 3X + 5 ) - 2 * ( X / 2 ) , Dimension of one side of the

board

--> D2 = ( 3X + 4 ) - 2 * ( X / 2 ) , Dimension of the other side of

the board

--> D1 = 3X + 5 - X , Dimension of one side of the board

--> D2 = 3X + 4 - X , Dimension of the other side of the board

--> Produced by MATCAL Program

--> D1 = 2X + 5 , Dimension of one side of the board

--> D2 = 2X + 4 , Dimension of the other side of the board

--> D1 = (2X + 5 ) m , Dimension of one side of the board

--> D2 = (2X + 4 ) m, Dimension of the other side of the board

--> Area of a rectangle is Length x Width i.e A = LW

--> Area of the rectangle is = 298 m^2 (square meter)

--> A = (2X + 5)(2X + 4)

--> 298 = (2X + 5)(2X + 4)

--> 298 = 2X * 2X + 2X * 4 + 5 * 2X + 5 * 4

--> Produced by MATCAL Program

--> 298 = 4X^2 + 8X + 10X + 20

--> 298 = 4X^2 + 18X + 20

--> Subtract 298 from both side of the equation

--> 298 - 298 = 4X^2 + 18X + 20 - 298

--> 0 = 4X^2 + 18X + 20 - 298

--> 0 = 4X^2 + 18X -278

--> F(X) = 4X^2 + 18X -278

The equation that represents the area of the board is given below as

--> F(X) = 4X^2 + 18X -278

--> (II)

--> Solve for the value of X using quadratic formula

--> The quadratic formula is X = (-b (+&-1) Sqrt(b^2 - 4ac) ) / 2a

--> a = 4 , b = 18 c = -278

--> X = - 18 + Sqrt(18^2 - 4 * 4 * -278) / 2 * 4

--> X = - 18 - Sqrt(18^2 - 4 * 4 * -278) / 2 * 4

--> X = - 18 + Sqrt(324 - 4 * -1112) / 2 * 4

--> X = - 18 - Sqrt(324 - 4 * -1112) / 8

--> Produced by MATCAL Program

--> X = - 18 + Sqrt(324 - 4 * -1112) / 8

--> X = - 18 - Sqrt(324 - 4 * -1112) / 8

--> X = - 18 + Sqrt(324 + 4448) / 8

--> X = - 18 - Sqrt(324 + 4448) / 8

--> X = - 18 + Sqrt(4772) / 8

--> X = - 18 - Sqrt(4772) / 8

--> X = - 18 + (69.0796641566822) / 8

--> X = - 18 - (69.0796641566822) / 8

--> X = - 18 + (69.08) / 8

--> X = - 18 - (69.08) / 8

--> X = 51.08 / 8

--> X = -87.08 / 8

--> X = 6.385

--> X = -10.885

--> Produced by MATCAL Program

--> The value of X = 6.39

--> The value of X = -10.89

--> An extraneous answer for the value of X is not acceptable,

therefore, the only

--> reasonable value for X must be greater than zero.

--> The acceptable value for X = 6.385

--> The height of the cuboid is X / 2. Therefore, the value of X

--> must be divided by 2 to get the actual value.

--> The height(H)of the cuboid = X / 2. Therefore, the value of X

--> The height(H)of the cuboid = 6.385 / 2

--> Produced by MATCAL Program

--> The height(H)of the cuboid = 3.1925

--> The height(H)of the cuboid = 3.19 m

--> D1 = 2X + 5 , Dimension of one side of the board

--> D2 = 2X + 4 , Dimension of the other side of the board

--> D1 = 2x 6.385 + 5 , Dimension of one side of the board

--> D2 = 2x 6.385 + 4 , Dimension of the other side of the board

--> D1 = 12.77 + 5 , Dimension of one side of the board

--> D2 = 12.77 + 4 , Dimension of the other side of the board

--> Produced by MATCAL Program

--> D1 = 17.77 , Dimension of one side of the board

--> D2 = 16.77 , Dimension of the other side of the board

--> D1 = 17.77m , Dimension of one side of the board

--> D2 = 16.77m , Dimension of the other side of the board

--> Length of the cuboid= 17.77 m , Dimension of one side of the board

--> Width of the cuboid= 16.77 m , Dimension of one side of the

board

--> Produced by MATCAL Program

--> (III)

--> Solving for the minimum point of the function(Vertex)

--> X = -b / 2a

--> a = 4 , b = 18 c = -278

--> X = -1 x 18 / 24

--> X = -36

--> X = -2.25

--> Solving for the point on the vertical co-ordinate by

substituting the value of X into F(X)

--> F(X) = 4X^2 + 18X -278

--> F(X) = 4 x -2.25^2 + 18 x -2.25 + -278

--> F(X) = 4 x 5.0625 + -40.5 + -278

--> F(X) = 20.25 + -40.5 + -278

--> F(X) = -298.25

--> The minimum point of F(X) exists/occurs at X = -2.25 Y =

-298.25

--> The coordinate of the vertex(minimum point) = ( -2.25 ,

-298.25 )

--> The extrema occurs at = ( -2.25 , -298.25 )

--> The function is concave up i.e the function has a minima

--> Produced by MATCAL Program

-->

--> (IV)

-->

--> Solving for the point of intersection of F(X) and DY/DX

--> Solving for the DY/DX

--> F(X) = 4X^2 + 18X -278

--> F'(X) = DY/DX = 2 x 4X + 18 x 1 + 0

--> F'(X) = DY/DX = 8X + 18

--> F'(X) = 8X + 18

--> Equate F(X) with DY/DX i.e set them with each other

--> 8X + 18 = 4X^2 + 18X -278

--> Combine like terms together using distributive property

--> 8X + 18 = 4X^2 + 18X -278

01-13- Every Time I Close My Eyes.mp3

--> 0 = 4X^2 + 18X -278 - 8X - 18

--> 0 = 4X^2 + 10X -296

--> The new Equation F(X) = 4X^2 + 10X -296

--> F(X) = 4X^2 + 10X -296

--> a = 4 , b = 10 c = -296

--> X = - 10 + Sqrt(10^2 - 4 * 4 * -296) / 2 * 4

--> X = - 10 - Sqrt(10^2 - 4 * 4 * -296) / 2 * 4

--> X = - 10 + Sqrt(100 - 4 * -1184) / 2 * 4

--> X = - 10 - Sqrt(100 - 4 * -1184) / 8

--> Produced by MATCAL Program

--> X = - 10 + Sqrt(100 - 4 * -1184) / 8

--> X = - 10 - Sqrt(100 - 4 * -1184) / 8

--> X = - 10 + Sqrt(100 + 4736) / 8

--> X = - 10 - Sqrt(100 + 4736) / 8

--> X = - 10 + Sqrt(4836) / 8

--> X = - 10 - Sqrt(4836) / 8

--> X = - 10 + (69.5413546028548) / 8

--> X = - 10 - (69.5413546028548) / 8

--> X = - 10 + (69.5413546028548) / 8

--> X = - 10 - (69.5413546028548) / 8

--> X = 59.5413546028548 / 8

--> X = -79.5413546028548 / 8

--> Produced by MATCAL Program

--> X = 7.44266932535685

--> X = -9.94266932535685

--> X = 7.44

--> X = -9.94

--> Both F(X) and DY/DX intersects at the value below

--> Points of intersection are X = 7.44 , X = -9.94

--> F(X) = 4 x 7.44266932535685^2 + 10 x

7.44266932535685 -296

--> F(X) = 221.573306746431 + 74.4266932535685 -296

--> F(X) = 296 -296

--> F(X) = 0.00

--> F(X) = DY/DX = 8X + 18

--> F(X) = DY/DX = 8 x 7.44266932535685 + 18

--> F(X) = DY/DX = 59.5413546028548 + 18

--> F(X) = DY/DX = 77.5413546028548

--> At X = 7.44266932535685 , F'(X) = DY/DX =

77.5413546028548

--> At X = 7.44 , F'(X) = DY/DX = 77.54

--> --------------------------------

--> F(X) = DY/DX = 4X^2 + 18X -278

--> F(X) = DY/DX = 4 x -9.94266932535685^2 + 18 x

-9.94266932535685 -278

--> F(X) = DY/DX = 4 x 98.8566733133921 +

-178.968047856423 -278

--> F(X) = DY/DX = 395.426693253569 + -178.968047856423

-278

--> F(X) = DY/DX = -61.5413546028548

--> F(X) = DY/DX = -61.54

--> At X = -9.94266932535685 , F(X) = DY/DX = -61.54

--> At X = -9.94 , F(X) = DY/DX = -61.54

--> The intersection point of F(X) and DY/DX are given below

--> At X = 7.44 , F'(X) = DY/DX = 77.54

--> At X = -9.94 , F(X) = DY/DX = -61.54

--> Produced by MATCAL Program

--> (V)

--> Volume = Length x Width x Height

--> Dimension 1 =17.77

--> Dimension 2 =16.77

--> Heigth =6.385

--> Volume = 17.77 x 16.77 x 6.385

--> Volume = 1902.7485165

--> Volume = 1,902.75

--> Volume = 1,902.75 m^3 (cubic metre)

--> Produced by MATCAL Program

--> Solve for the volume using the given height

--> Given height = 3.5

--> D1 = (2X + 5 ) m , Dimension of one side of the board

--> D2 = (2X + 4 ) m, Dimension of the other side of the board

--> D1 = (7 + 5 ) m , Dimension of one side of the board

--> D2 = (7 + 4 ) m, Dimension of the other side of the board

--> D1 = 12 , Dimension of one side of the board

--> D2 = 11 m, Dimension of the other side of the board

--> D1 = 12 , Dimension of one side of the board

--> D2 = 11 m, Dimension of the other side of the board

--> Volume = 462 m, Dimension of the other side of the board

--> Volume = 462.00 m, Dimension of the other side of the board

--> Area = Length x Width

--> Area = 12 x 11

--> Area = 132

--> Area = 132.00 m^2 (square meter)

--> (VI)

--> Range = [ -298.25 , + Inf ]

--> Range : F(X) = Y >= -298.25

--> End of Quad 1

--> Produced by MATCAL Program

The equation that represents the area of the board is given below as

--> F(X) = 4X^2 + 18X -278

--> Proceed to check if the value of X obtain is correct/accurate.

--> X = 6.39 , -10.89 The value of X is correct because the

equation is equal to zero.

--> F(X) = 4 x 6.385^2 + 18 x 6.385-278

--> F(X) = 4 x 40.768225 + 18 x 6.385-278

--> F(X) = 4 x 40.768225 + 114.93-278

--> F(X) = 163.0729 + 114.93-278

--> F(X) = 278.0029-278

--> F(X) = 2.89999999997193E-03

--> F(X) = 0.00

--> F(X)------- The value of X is correct because the equation is

equal to zero.

--> F(X) = 4 x -10.885^2 + 18 x -10.885-278

--> F(X) = 4 x 118.483225 + 18 x -10.885-278

--> F(X) = 4 x 118.483225 + -195.93-278

--> F(X) = 473.9329 + -195.93-278

--> F(X) = 278.0029-278

--> F(X) = 2.89999999996482E-03

--> F(X) = 0.00

--> F(X)------- The value of X is correct because the equation is

equal to zero.

--> F(X)------- If the F(X) is not equal to zero, then the value of X

is not correct.

--> F(X)------- However, F(X) is equal to zero which proves that

the value of X is correct/accurate.

--> Produced by MATCAL Program

01-01- The Moment (Album Version).mp3