Find the dimensions of the rectangle of greatest area that can be inscribed in a semicircle of radius?

Question

please, help

h=2^1/2 , i=2X(2)^1/2

Answer 1

ill give u a hint
http://i55.photobucket.com/albums/g149/delta_bravo_1kilo/yahoo/2circles.jpg

Answer 2

To find the rectangle of largest area in a semicircle, first find the rectangle of largest area in a circle.

We find that the rectangle with the largest area in a circle is actually a square.

let square have side 'a' and circle have radius 'r'.
Since the two diagonally opposite points of the square form the diameter of the circle, we have
2r = sqrt(2)*a
ie, a = sqrt(2)*r

now, draw a line that divides the square into 2 by picking the mid-points of opposite sides of the square. Extent these lines to the boundary of the circle. Each half of the figure now formed has a rectancle of largest area inscribed in a semicircle of radius r.

the dimensions are
length = sqrt(2)*r [ since one side of the square is retained ]
height = r/sqrt(2) [ since the opposite sides are cut in half ]

Answer 3

let the length and breadth of the rectangle be l and b, A be area=lb.
since the angles of a rectangle are 90 and angle in a semicircle is also 90, so the diagonal of rectangle are actually diameter of circle. By pythagorus therom
l^2 + b^2=4r^2
put l=A/b
A^2/b^2 + b^2=4r^2
=>A=sqrt(b^2(4r^2-b^2)) .............1
to maximize A, put dA/db=0
(-2b^2)/sqrt(4r^2-b^2) +sqrt(4r^2-b^2 )=0
=>-2b^2+4r^2-b^2=0
=>b=sqrt(4/3)r
now A=sqrt(32/9)r^2 by putting value of b in terms of r frm 1
A=bl
=>l=A/b
=>l=sqrt(8/3)r and b=sqrt(4/3)r are the dimension of rectangle of largest area that can be inscribed in a circle of radius r


for a semicircle
the rectangle of largest area will be symitrical abt the radius which divides semicircle in half.
let height of rectangle be b and length be l, area A
one of l lie along the diameter.
by phythagorus th.
b^2+l^2/4=r^2
again put l=A/b
and find A
A=sqrt(4b^2(r^2-b^2))
find dA/db and put this=0 and find b in erms of r
b=r/sqrt(3)
=>A=r^2sqrt(8/9)
=>l=sqrt(8/3)r b=sqrt(1/3)r

Answer 4

Let the radius be r
Then area = 2rsinxrcosx = 2r^2sincosx , where x is the angle the radius makes with the upper right vertex of the rectangle.
This simplifies to A = r^2sin(2x)

Now take the derivative dA/dx, set it to 0 and solve for x.

Use the value of x found to fin max area.

Answer 5

You need to give the radius