First plot this function in the first quadrant.

`y = (4-x)/(2+x)`

Now if you plot this correctly (I plot this on MATLAB), you will see that its curve which has y-intercept at y = 2 and x-intercept at x =4.

The rectangle A(x) varies within the enclosed area by this curve and limited by the curve, y-axis and x-axis.

Therefore, The width of rectangle is `x` .

The height of rectangle is `y = (4-x)/(2+x)`

Therefore, A(x) is,

`A(x) = x*y`

`A(x) = x*((4-x)/(2+x))`

`A(1) = 1*((4-1)/(2+1))`

`A(1) = 1*(3/3) = 1`

**A(1) = 1.**

The maximum area can be found by differentiating A(x) wrt x.

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

`(dA)/(dx) = (((4-2x)*(2+x))-((4x-x^2)*1))/(2+x)^2`

`(dA)/(dx) = (2(2-x)*(2+x)-(4x-x^2))/(2+x)^2`

`(dA)/(dx) = (8-2x^2-(4x-x^2))/(2+x)^2`

`(dA)/(dx) = -(x^2+4x-8)/(2+x)^2`

`(dA)/(dx) = -((x-(-2+2sqrt(3)))(x-(-2-2sqrt(3))))/(2+x)^2`

For extreme points (dA)/(dt) = 0

Then answers are `x = -2+2sqrt(3)` or `x = -2-2sqrt(3)` .

But x cannot be negative, so,

`x = -2+2sqrt(3) = 1.4641` which is a valid answer since it lies within 0 and 4.

If this is a maximum, the sign of the second derivative just be negative.

`(dA)/(dx) = -(x^2+4x-8)/(2+x)^2`

`(d^2A)/(dx^2) = -[((2x+4)(2+x)^2-(x^2+4x-8)2(2+x))/(2+x)^4]`

`(d^2A)/(dx^2) = -[(2(2+x)^2-2(x^2+4x-8))/(2+x)^3]`

`(d^2A)/(dx^2) = -2[((2+x)^2-(x^2+4x-8))/(2+x)^3]`

`(d^2A)/(dx^2) = -2[(x^2+4x+4-(x^2+4x-8))/(2+x)^3]`

`(d^2A)/(dx^2) = -2[12/(2+x)^3]`

`(d^2A)/(dx^2) = -[24/(2+x)^3]`

For x>0, `(d^2A)/(dx^2) lt 0` (negative).

Therefore at `x = -2+2sqrt(3)` we have a maximum for A.

The maximum value of A is,

`A(-2+2sqrt(3)) = (-2+2sqrt(3))*((4-(-2+2sqrt(3)))/(2+(-2+2sqrt(3))))`

`A(-2+2sqrt(3)) = (-2+2sqrt(3))*((6-2sqrt(3)))/(2sqrt(3))`

`A(-2+2sqrt(3)) = 2(sqrt(3)-1)*((3-sqrt(3)))/(sqrt(3))`

`A(-2+2sqrt(3)) = 2(sqrt(3)-1)*(sqrt(3)-1))`

`A(-2+2sqrt(3)) = 2(sqrt(3)-1)^2`

**The maximum area of the rectangle is** `2(sqrt(3)-1)^2`

**Decimal approximation of answer is 1.07179677.**