Find the maximum area of a rectangle inscribed in a circle of radius 12 in?

Answer 1

The length and width of the rectangle (and thus it's area) will be defined by the circle. Specifically, the diagonal of that rectangle will always be equal to the diameter of the circle.

To find the maximum area then, one can simply define the rectangle with that relationship, take the derivative of that function, and find out where that comes to a value of zero. The area will be peaked at that point:

a = lw

d2 = l2 + w2

∴ l = (d2 - w2)1/2

Remember that the radius is 12, so the diameter is 24, and things can be simplified by plugging that into the equation at this point:

l = (576 - w2)1/2

∴ a = (576 - w2)1/2 w

The next step is to take this equation for area, and find it's rate of change with respect to width:

∴ da/dw = (576 - w2)1/2 + 1/2 * (576 - w2)-1/2 * -2w * w

∴ da/dw = (576 - w2)1/2 - (576 - w2)-1/2 * w2

Now let that value equal zero:

0 = (576 - w2)1/2 - (576 - w2)-1/2 * w2

∴ (576 - w2)1/2 = (576 - w2)-1/2 * w2

∴ 576 - w2 = w2

∴ w2 = 576 / 2

∴ w2 = 288

∴ w ≈ 16.97

And one can work out the corresponding height:

d2 = l2 + w2

∴ 242 = l2 + 288

∴ l2 = 576 - 288

∴ l2 = 288

∴ l ≈ 16.97

Meaning that the optimum rectangle is a perfect square.

To be completely thorough, it should be confirmed that the area found was a maximum and not a minimum. This can be done easily enough by taking a slightly lesser width and a slightly greater width, and seeing how their areas compare:

a = (576 - w2)1/2 w

Let w = 16

∴ a = (576 - (16)2)1/2 * 16

∴ a = 3201/2 * 16

∴ a = 51/2 * 128

∴ a ≈ 286.22

Let w = 18

a = (576 - 182)1/2 * 18

∴ a = 2521/2 * 18

∴ a = 71/2 * 108

∴ a ≈ 285.74

Both of these values are less than the area surrounded by the square, so the square's area is indeed a maximum and not a minimum.