Question

A rectangle with its base on the x-axis has its upper vertices on the curve of y=3-x^2, so find the maximum perimeter of such a rectangle.

Answer 1

erm...
all I got so far is that domain restriction is 3-x^2=0

Answer 2

@sourwing

Answer 3

ok, I got it :)
had to use this to get the right idea tho xD
http://www.algebra.com/algebra/homework/Surface-area/Surface-area.faq.question.310349.html

so from that we get that the two points on the parabola that would make up the vertices of the rectangle
(x, 3-x^2) and (-x, 3-x^2)
so L=2x
and W=3-x^2

Answer 4

then Perimeter is
P = 2L + 2W plug in
= 2(2x) + 2(3-x^2)
= 4x + 6 - 2x^2 <- so take derivative of that

Answer 5

one sec 314

Answer 6

-4x+4?

Answer 7

yup :)
then solve for x
for -4x + 4 = 0

Answer 8

1

Answer 9

yep :)
then plug x = 1 back into
L = 2x
W = 3-x^2

then into P = 2L + 2W

Answer 10

l = 2(1) = 2
w = 3-1^2 = 3
p=2(2) + 2(3) = 4 + 6 = 10

Answer 11

erm
w = 3-1^2 = 3 - 1 = ?

Answer 12

2?

Answer 13

I think so xD

Answer 14

so 2 is the answer righT?

Answer 15

mmm na, I was just correcting your "w = 3-1^2 = 3"

Answer 16

alright, so where i am goin' from here?

Answer 17

l = 2(1) = 2
w = 3-1^2 = 2
p=2(2) + 2(2) = 4 + 4 = 8 <- tah-dah! XD

Answer 18

is it's 8?

Answer 19

<3

Answer 20

yeah, I think so :)

Answer 21

yep