calculus!!!a rectangle is to be inscribed in a semicircle of radius 8, with one side lying on the diameter of?

Question

A rectangle is to be inscribed in a semicircle of radius 8, with one side lying on the diameter of the circle. What is the maximum possible area of the rectangle? will you please explain how to do this??

Answer 1

Geometry problem.
A square (a unique "rectangle").
x = (64/5) ^(1/2),
y = 2*x = 2*(64/5)^1/2
Max. Area = 2*x*y = 4*64/5 = 51.2.

Check by substituting x & y in circle equation.

Answer 2

the utmost possible section is whilst the width of rectangle is a million/2 of its lenght.in case you draw the radius of semicircle it relatively is wide-spread to diameter and divide it to 2 smaller rectangle and likewise assume the lenght it relatively is positioned on diameter is 2x and the width is y, then you definitely will see the a million/2 rectangle diameter(surely the radius of circle) is 8 and we've: x^2+y^2=8^2=sixty 4. this section would be maximum if x=y So x^2+y^2=2x^2=sixty 4 or x^2=32 section=2xy=2x^2=2(32)=sixty 4

Answer 3

Let the origin of a coordinate system be the center of the semicircle.
Define θ as a positive angle from the y-axis.
Then
A = 2*64sinθcosθ
dA/dθ = 128(cos^2θ - sin^2θ) = 0 for max area
(cosθ + sinθ)(cosθ - sinθ) = 0
θ as defined may not exceed 90°, so
sinθ = cosθ
tanθ = 1
θ = 45°
sin45°cos45° = 0.5
A = 2*64/2 = 64 square units

Check:
sin46°cos46° = 0.499695413509547865003121
sin44°cos44° = 0.499695413509547865003121

Answer 4

Circle:
x^2 + y^2 = 64
y^2 = 64 - x^2
y = sqrt(64 - x^2) is a semi-circle
~~
Area = length * width
A(x) = (2x) * ( f(x) )
A(x) = 2x*sqrt(64 - x^2)
A(x) = 2sqrt(64x^2 - x^4)
~~
u = 64x^2 - x^4
du = 128x - 4x^3 dx
du/dx = 128x - 4x^3
~~
A(x) = 2sqrt(u)
A(x)/du = 1/sqrt(u)
A(x)/du = 1/sqrt(64x^2 - x^4)
~~
A(x)/du * du/dx = A'(x)
~~
A'(x) = (128x - 4x^3)/sqrt(64x^2 - x^4)
A'(x) = (128 - 4x^2)/sqrt(64 - x^2)
0 = (128 - 4x^2)/sqrt(64 - x^2)
0 = 128 - 4x^2
4x^2 = 128
x^2 = 32
x = +-sqrt(32)
Rectangle with length and width of sqrt(32) units.