Question

Help Please Please Please! (:
Optimization: A rectangle has one side on the x-axis and two vertices on the curve y=3 / x^2+2. Find the vertices of the rectangle with maximum area.

Answer 1

not nearly as hard as it looks
base will be \(2x\) and height will be \(\frac{3}{x^2+2}\) therefore the area of the rectangle will be
\[A(x)=\frac{6x}{x^2+2}\]
maximize that one

Answer 2

I find A'(x) right? @satellite73

Answer 3

\[A'(x)= \frac{ 12x }{ x^2+2 } - \frac{ 12x^3 }{ (x^2+2)^2 }\]

Answer 4

\[A'(x)=0\]
x=0
@satellite73

Answer 5

I got
\( A'(x) = \dfrac{-12x^2} {(x^2 + 2)^2} + \dfrac{6}{x^2 + 2} \)

Answer 6

It can be simplified to:

\(A'(x) = \dfrac{-6x^2 + 12}{(x^2 + 2)^2} \)

Answer 7

how'd you get a negative 12?

Answer 8

nevermind I caught my mistake.. @mathstudent55

Answer 9

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

Answer 10

right? @mathstudent55

Answer 11

correct

Answer 12

So where so I go from there? @mathstudent55

Answer 13

Now plug in those two values of x in the original function to find the corresponding y-coordinates of the two points on the function.

Answer 14

you are done

Answer 15

the original finction of y= or a(x)=?

Answer 16

\(A(x)\) will give you the maximum area, but you are asked for the vertices

Answer 17

I don't understand.

Answer 18

so they are \((-\sqrt2,0)\) and \((\sqrt2,0)\) for the ones on the \(x\) axis

Answer 19

The function.