Asked by Mable
A rectangle is inscribed into a semi circle at radius 2. What is the largest area it can have and what are the dimensions
Answers
Area= 4 max
base =2sqrt2
height = sqrt2
Help is always appreciated :)
Answers
Area= 4 max
base =2sqrt2
height = sqrt2
Help is always appreciated :)
Answers
Answered by
Steve
base = 2x
height = y = √(4-x^2)
a = 2x√(4-x^2)
da/dx = 4(2-x^2)/√(4-x^2)
da/dx=0 when x = √2
so, max area is 2(√2)√(4-2) = 4
height = y = √(4-x^2)
a = 2x√(4-x^2)
da/dx = 4(2-x^2)/√(4-x^2)
da/dx=0 when x = √2
so, max area is 2(√2)√(4-2) = 4
Answered by
Anonymous
I will work on 1/4 of the circle
y = height
x = half of width
r^2 = x^2 + y^2 = 4
A = x y so x = A/y
A^2/y^2 + y^2 = 4
A^2 + y^4 = 4 y^2
let p=y^2
p^2 -4p = -A^2
p^2 - 4p + 4 = -A^2+4
(p-2)^2 = - (A^2-4)
p = 2
so
y = sqrt 2 and A = 2
and x = 2/sqrt2 = sqrt 2
so
height= y = sqrt 2
width = 2 x = 2 sqrt 2
A = 4
y = height
x = half of width
r^2 = x^2 + y^2 = 4
A = x y so x = A/y
A^2/y^2 + y^2 = 4
A^2 + y^4 = 4 y^2
let p=y^2
p^2 -4p = -A^2
p^2 - 4p + 4 = -A^2+4
(p-2)^2 = - (A^2-4)
p = 2
so
y = sqrt 2 and A = 2
and x = 2/sqrt2 = sqrt 2
so
height= y = sqrt 2
width = 2 x = 2 sqrt 2
A = 4
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