Find the dimensions of the rectangle of greatest area which can be inscribed in a circle of radius 1.
2007-04-10 16:11:38 · 3 answers · asked by bebe_love_xo 1 in Science & Mathematics ➔ Mathematics
A square with a radius of 1, a side of SQRT(2) and an area of 2.
2007-04-10 16:20:14 · answer #1 · answered by suesysgoddess 6 · 1⤊ 0⤋
Why will it be a square with sides = √2?
A circle with radius 1 centered at the origin has the equation
x^2 + y^2 = 1
Any inscribed rectangle will have an area
A = 4xy for a point (x,y) in the first quadrant.
y = A/4x
x^2 + (A/4x)^2 = 1
16x^4 + A^2 = 16x^2
16x^4 - 16x^2 + A^2 = 0
Since you said calculus,
64x^3 - 32x = 0 for max area
x(2x^2 - 1) = 0
x = 0 is a minimum, and trivial
(2x^2 - 1) = 0 is a difference-of-squares
(x√2 + 1)(x√2 - 1) = 0
x = -1/√2,1/√2
We're looking for the positive value, so
x = 1/√2, which is half the horizontal length, 2/√2
Substituting 1/√2 for x in x^2 + y^2 = 1 we have
(1/√2)^2 + y^2 = 1
1/2 + y^2 = 1
y^2 = 1/2
y = 1/√2, which is half the vertical dimension, 2/√2
Rationalizing these dimensions we have
L = 2√2/(√2√2) = 2√2/2 = √2
W = √2
so the rectangle is a square with side √2
2007-04-10 17:09:14 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
it will be a square with a diaganol line of 2 and an area of 2 also
2007-04-10 16:15:50 · answer #3 · answered by pghpanthers2 2 · 0⤊ 0⤋