What is the maximum area of a rectangle inscribed in a given region?

[skateza]

Homework Statement

Find the area of the largest rectangle that can be inscribed in the region bounded by the graph of y = (4-x)/(2+x) and the coordinate aces in the first quadrant.


I think my only problem with this one is taking the derivative,

this is what i get y' = (-x^2 - 4x + 8)/(2+x)^2

Critical numbers: [1-root(48)]/-2, but that doesn't seem to be giving me a maximum value, can someone take a second look this.

 

[BlackWyvern]

If I'm not mistaken:

A = xy, x = x, y = (4-x)/(2+x)

A = x(4-x/2+x)
= (4x - x^2 / 2 + x)

Using Quotient Rule (feel free to use the product rule if you want to):
dy/dx = ((2 + x)(4 -2x) - (4x - x^2)(1)) / (2 + x)^2

=8 -4x +4x -2x^2 - 4x + x^2 / g^2
dy/dx = 8 -x^2 -4x / 4 + 4x + x^2

Graphing that, I can see 2 roots.
(can't be bothered actually solving properly for them now though, :) )
They are:
-5.46410 & 1.464101

Using 1.464101 leads to an area of about 1.0717
Graphing the original function to be optimized, I see that this is correct.

 

[BlackWyvern]

Okay, doing a poly long divide, I get the dy/dx as:
-1 + 12/(x+2)(x+2) = 0
==>
x^2 + 4x - 8 = 0

This is interestingly the numerator of the original function.

Solving this with quadratic formula yields the same results.

 

[skateza]

my error was in using the quadratic formula, i took -a instead of -b which is why i got a strange value.