# Given the rectangle whose top corners are on the curve x^2 = 1 - y and base is on x axis, calculate how large area of rectangle could be?

### 1 Answer | Add Yours

You need to use the equation that gives the area of rectangle, such that:

`A = L*w`

`L` represents the length of rectangle (measured on x axis)

`w` represents the width of rectangle (measured on y axis)

Since the problem provides the information that the top corners of rectangle lie on the curve `x^2 = 1 - y` , you need to re-write the equation of the curve in terms of x, such that:

`y = 1 - x^2 => w = 1 - x^2`

Since the base of rectangle is found on x axis, yields that that `L = 2x`

Replacing `2x` for L and `1 - x^2` for w yields:

`A(x) = 2x(1 - x^2) => A(x) = 2x - 2x^3`

Since the problem requests for you to evaluate the maximum area of rectangle, you need to differentiate the equation of area with respect to x, such that:

`A'(x) = 2 - 6x^2`

The area of rectangle is maximum if `A'(x) = 0` , such that:

`A'(x) = 0 => 2 - 6x^2 = 0 => -6x^2 = -2 => x^2 = 2/6 => x^2 = 1/3 => x_(1,2) = +-1/sqrt 3`

Since x cannot be negative, you need to keep the positive value `x = 1/sqrt3` .

You need to evaluate y = w, such that:

`w = 1 - x^2 = 1 - 1/3 = 2/3`

`A = 2*1/sqrt3(2/3) => A = 4/(3sqrt3) => A = (4sqrt3)/9`

**Hence, evaluating the largest area of the given rectangle, under the given conditions, yields **`A = (4sqrt3)/9.`