ANSWERS EXPECTED FROM VERIFIED MEMBERS!
Given that there exist an rectangle between the x-axis and the parabolic curve y=8-x² with one side coinciding with x axis and other 2 points touching the curve. Then find the area of the biggest possible rectangle that can satisfy the conditions?
®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®®
Answers
Answered by
3
Hope u understand my writing
Attachments:
07161020:
thanks
I m ryt?
Wrong it seems to me.
how
what's the method
Seems ok. Sry
no u r ryt
I make a mistake there that it is 2xy instead of 4xy
but u can get answer by divide it by 2
ok
Answered by
3
See the rough sketch of the parabola y = 8 - x^2 in the diagram enclosed.
ABCD is the rectangle symmetrically positioned about y axis. It's symmetric as the parabola is symmetric.
Area (ABCD) = A = 2 x y . This is to be maximized.
A = 2 x (8 - x^2) = 16 x - 2 x^3
Differentiate and Let dA/dx = 0.
A' = 16 - 6 x^2 = 0.
x = + or - 2 * sqrt(2/3).
y = + 16/3.
So the rectangle is formed by A(- sqrt (83), 0), B ( sqrt(8/3), 0), C(sqrt(8/3), 16/3), D(- sqrt(8/3), 16/3).
Max Area = 2 x y = 64/3 * sqrt(2/3) units.
ABCD is the rectangle symmetrically positioned about y axis. It's symmetric as the parabola is symmetric.
Area (ABCD) = A = 2 x y . This is to be maximized.
A = 2 x (8 - x^2) = 16 x - 2 x^3
Differentiate and Let dA/dx = 0.
A' = 16 - 6 x^2 = 0.
x = + or - 2 * sqrt(2/3).
y = + 16/3.
So the rectangle is formed by A(- sqrt (83), 0), B ( sqrt(8/3), 0), C(sqrt(8/3), 16/3), D(- sqrt(8/3), 16/3).
Max Area = 2 x y = 64/3 * sqrt(2/3) units.
Attachments:
:-) :-)
area is 4xy n
how 4xy?
it will be 2xy
answer is correct murty sir!!!
thanks
Best answer sir !
Similar questions