What is the area of the largest rectangle with its base on the x-axis and the vertices above the x-axis and on the curve y=4-2x^2?

Question

I've plotted the points and see that its a downward facing parabola with a global maximun at (0,40. I am confused as to how I can apply calculus to get the area of the rectangle. I am thinking this is a limit question, antiderivative or inflection point type of question. Any insight would be deeply appreciated!

Answer 1

You mean global maximum at (0,4)---right?

If two corners of the rectangle are on that curve, it means they have the same y value. The basic formula of the parabola is y = x² , which means there are 2 x values for a range of y values.

Let's define the dimensions of the rectangle:

The parabola is centered on the y axis, and the x value is plus or minus, so the horizontal dimension of the rectangle is 2x

The vertical dimension is simply y, which is defined as 4 - 2x² . Therefore the area of the rectangle is:

A = 4x - 2x³

To find the maximum A, take the 1st derivative and set = to 0:

A' = 4 - 6x² = 0, and x = √(2/3) = 0.816

Answer 2

The graph of y = 4 - 2x² is a parabola opening downward with vertex at (0,4).

A rectangle with base along the x-axis and upper vertices on the parabola is symmetric to the y-axis with base 2x and height 4 - 2x², where x is the distance from the origin to the lower right corner of the rectangle.

Area of rectangle = A = 2x(4 - 2x²), 0 < x < √2

A = 8x - 4x³

dA/dx = 8 - 12x² = 0 when x² = 2/3.

So, x = √(2/3) ≈ 0.816

When 0 < x < √(2/3), A' > 0. So, A is increasing.

When √(2/3) < x < √2, A' < 0, So, A is decreasing

Therefore, there is a relative and absolute maximum when x = √(2/3) and y = 4 - 2(2/3) = 8/3

Maximum area = 2√(2/3)(8/3) = (16/3)√(2/3) ≈ 4.35