Question

what is the max perimeter all all rectangles that can be inscribed in

x^2/a^2 + y^2/b^2 = 1

Answer 1

am trying to type, software glitchy

Answer 2

so, we have to maximize 2(x+y)
solve for y in x^2/a^2 + y^2/b^2 = 1
then put it back in
f(x)=x+....
then 'x' in f'(x)=0

Answer 3

u know how to maximize, right ?

Answer 4

Why maximize 2(x + y)? The perimeter is 2(a + b) isnt it?

Answer 5

you'll get x and y in terms of a and b
then you put those values in P =2(x+y) and you get perimeter in terms of a and b.
and its not 2(a+b)

Answer 6

oh ya sorry, i swapped the variables on ellipse

y = b/a sqrt(a^2 - x^2)
y' = -(b x)/(a sqrt(a^2-x^2))

p = 2(x + y)
p' = 2(1 + y')

p' = 0 whenever y' = -1

y' = -(b x)/(a sqrt(a^2-x^2)) = -1

x = a^2/sqrt(a^2 + b^2)
y = b/a sqrt(a^2 - [a^2/sqrt(a^2 + b^2)]^2)

x + y = sqrt(a^2+b^2)
2(x + y) = 2 sqrt(a^2 + b^2)

but (sorry glitchy google chrome/openstudy)

Answer 7

but

Answer 8

2(x+y) only gives the max perimeter of upper portion of ellipse , would it not be 4(x+y)

Answer 9

i solve for x, but

Answer 10

nevermind, got it.

Answer 11

you mean to say, that you just took, positive value of 'y', thats why the 'upper' part ?

Answer 12

ya , for the upper region of the ellipse, i see thank you you are clever

Answer 13

changing browsers though never use chrome here

Answer 14

and since you get the same thing when you do it for negativevalue of 'y', you say,its 2+2=4 ?
i don't think thats the case, you take positive value of y because perimeter can't be negative.
i still think its 2(x+y) only.

Answer 15

what do you obtain ?

but y' = -1 implies

x = +- a^2/(a^2 + b^2)
y = +- b/a sqrt(a^2 - x^2)

i cannot ignore the sign can i ?
perimeter cannot be negative but y values can be. perimeters would be just absolute value of it

Answer 16

x=+- ....
so if you addd those 'x' values, you get 0 :P
and theres no reason to subtract.....
i don't have words to explain properly, but it'll be 2(x+y) only.