Question

If i have 36 = 8x^2

divide by 8 on both sides

I get 4.5 = x^2

How do I get rid of the x^2?

Answer 1

naturla log

Answer 2

I dont know what that is. Can you show me please?

Answer 3

answer would be 2.12

Answer 4

natural log might be bit too advanced
here is a nother way

Answer 5

\[\sqrt[2]{4.5}\]

Answer 6

x=√(5/2)

Answer 7

dont for got medal

Answer 8

dang, thats what i thought. now my whole problem looks even more screwy. agh!

Answer 9

You need to do this: \[(4.5)^{1/2}=(x ^{2})^{1/2}\]

This will cut the square amd you will have \[\sqrt[2]{4.5} = x\]

Answer 10

\[36 = 8x^2\]

\[\frac{36}{8}=x^2\]
\[\frac{9}{2}=x^2\]

\[x=\pm\sqrt{\frac{9}{2}}\]

Answer 11

You could take the root of 9. Still not very pretty...

Answer 12

one could simplify it to

\[x=\pm\frac{3\sqrt{2}}{2}\]

Answer 13

See now when I plug that into the original equation, im lost:

4x^2 + 9y^2 = 36
4(2.12) = 9Y^2 = 36
8.48 + 9y^2 = 36

Answer 14

I feel like this cant be right?

Answer 15

what did they want you to do with the original equation?

Answer 16

find the dimensions of a rectangle with the greatest area that can be inscrubed in the ellipse

Answer 17

i folowed all the steps, and i still want to shoot my brains out. :(

Answer 18

lol...don't do that

Answer 19

Im too cute for this crap, anyone want to support me for life? *sigh*... jk

Answer 20

What are the din=mentions of your rectangle in terms of x,y?

Answer 21

the center of the rectangle will be the center of the ellipse


then let x,y be the coordinate of the intersection of the ellipse and the rectangle in the first quadrent

Answer 22

then the area of the rectangle is

\[A=2x\cdot 2y\]

Answer 23

zarkon can you help me wiht my question

Answer 24

zarkon can you help me wiht my question

Answer 25

you want to maximize this subject to the constraint \[4x^2+9y^2=36\]

Answer 26

i poseted it

Answer 27

i'll look in a few mins

Answer 28

ok my notes say: solve for y, so i got y=sqrt 36-4x^2 / 3

Answer 29

\[y=\frac{\sqrt{36-4x^2}}{3}\]

yes

Answer 30

yeah, exactly that

Answer 31

then i get = 1/3x * sqrt 36-4x^2 to get rid of the denominator

Answer 32

?

Answer 33

thats how he did it in all the problems, we do the product rule next. wrong?

Answer 34

go back to \[A=2x\cdot 2y\]

and replace \[y\] with what you found it to be equal to.

Answer 35

\[A=2x\cdot 2\frac{\sqrt{36-4x^2}}{3}\]

Answer 36

now this is a function of one variable.

take the derivative ...set it to zero and solve for x

Answer 37

ok im gonna try that, then im going to shame my brain for being so dumb. Thank you!!

Answer 38

np ;)

Answer 39

Hmmm, I think you got it right, especially if you use the form
\[x=\frac{3\sqrt{2}}{2}\]
which is exact, rather than 2.12 which is approximate

Answer 40

Of course, you now must solve for y. and then the area of the rectangle is 4xy

Answer 41

As a check, the area of an ellipse is
\[Area= \pi a b\]
which for a=3 and b=2, gives about 18.85. The rectangle must have less area than this.

Answer 42

I get \[x=\frac{3\sqrt{2}}{2}\] when I maximize the area :)

Answer 43

Looks very familiar!