# Maximize the area of the rectangles inscribed the unit circle

*print*Print*list*Cite

### 2 Answers

You should notice that the diagonal of rectangle coincides with the diameter of the circumscribed unit circle, hence, since the radius measure 1 unit, the diameter measures 2 units.

You should come up with the notation for the sides of rectangle such that: w=width and l=length.

You need to use Pythagora's theorem in the right triangle whose hypotenuse is the diagonal of rectangle such that:

`d^2 = l^2 + w^2`

`4 = l^2 + w^2 =gt w^2 = 4 - l^2 =gt w = sqrt(4 - l^2)`

You need to evaluate the area of rectangle such that:

`A = l*w `

You need to write the area in terms of l such that:

`A(l) = l*sqrt(4 - l^2)`

You need to optimize the area of rectangle, hence, you need to differentiate the area with respect to l such that:

`A'(l) = l'*sqrt(4 - l^2) + l*(sqrt(4 - l^2))'`

`A'(l) = sqrt(4 - l^2) + (l(4-l^2)')/(2sqrt(4-l^2))`

`A'(l) = sqrt(4 - l^2) + (-2l^2)/(2sqrt(4-l^2))`

`A'(l) = sqrt(4 - l^2) + (-l^2)/(sqrt(4-l^2))`

You need to solve for l the equation `A'(l) = 0` such that:

`sqrt(4 - l^2) + (-l^2)/(sqrt(4-l^2)) = 0`

`(4 - l^2 - l^2)/(sqrt(4-l^2)) = 0`

Since `(sqrt(4-l^2)) != 0 =gt 4 - 2l^2 = 0 =gt 2l^2 = 4 =gt l^2 = 2 =gt l = sqrt2`

`w = sqrt(4 - 2) = sqrt2`

**Hence, the area of rectangle is maximum, A = 2, when the lengths of sides are equal such that `l=w=sqrt2` .**

**Sources:**

How many rectangles? If it is one rectangle, then the rectangle must be a square with square root of two for a side, so he area would be two