rectangle qustion

Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=6-x^2 . What are the dimensions of such a rectangle with the greatest possible area?

Find the width and Height.

Answer 1

let vertex of rectangle on x-axis is (-a,0) and (a,0), another words length of rectangle is 2a and width

is y = 6 - a². So, area A = 2a(6 - a²) = - 2a³ + 12a.

Now derivative dA/da = - 6a² + 12 = 0; a² =2; a = ± √2.

If a = √2 then because A'' = - 12a and A''(√2) = - 12√2 < 0 we have maximum area.

Length is 2√2 and width is 6 - 2 = 4.

SAo, maximum area A_max = 4·2√2 = 8√2

Answer 2

Since this is symmetrical, we can take twice the Area from the origin to a point on the positive x-axis.

Area, A(x) = base×height = 2xf(x)=2x(6-x²) = 12x-2x³

A'(x) = 12-6x²

Setting to zero and solving for x

x = √2

height = f(x) = 6-(√2)² = 4

Base of the rectangle = 2x = 2√2

Height of the rectangle = 4

Maximum Area = 2√2(4) = 8√2

Answer 3

This works best with a sketch of the curve and the inscribed rectangle. Please see video.