PLEASE I JUST NEED THE BY HAND CALCULATION OF FUNCTIONS USING NEWTONS METHOD, INTERPOLATION, OR BISECTION

Assume that the floor is the x$-$axis and the y$-$axis is at the centre of the door

$($see figure above$).$

The arcade is given by the polynomial function.

arc$()=0$

$81$

$62$

$43$

$2+0+1+2+3$

where the coefficients, $0,1,...$ are the digits of your student ID$.($exp:

ID$=153310,$ then $()=$

$85$

$63$

$43$

$2+12)$

Required work:

$$ You need to find the coordinate of the contact Arcade$-$Door points, such

that the door is the largest possible $($in term of surface, the maximum

surface$).$ Note that the problem is symmetric, you need to find only one

contact point $($between $0$ and $1)$ and to other one is simply the negative

value.

$$ You need to consider $4$ cases: $2$ rectangle doors and two square shaped

doors.

$$ For each case you change the arcade equation by considering the ID

digits of each group member. If you are less then $4$ students in the

group just permute the digits.

$$ You need to plot in your OCTAVE$/$MATLAB code the arcade and the

door $($use the plotting command for the doors: Rectangle $($position$,$

$[$x$\_$pos, y$\_$pos, door$\_$width, door$\_$hight$],$linewidth$,$ value$)$

x$\_$pos, y$\_$pos: are the left$-$bottom position of the door.Assume that the floor is the $x-$axis and the $y-$axis is at the centre of the door

$($see figure above$).$

The arcade is given by the polynomial function.

arc$(x)=-a_0{x}^8-a_1{x}^6-a_2{x}^4-a_3{x}^2+a_0+a_1+a_2+a_3$

where the coefficients, $a_0,a_1,$dots are the digits of your student ID$.($exp:

ID $=153310,$ then $\{$:arc$(x)=-x^8-5x^6-3x^4-3x^2+12)$