Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=2−x2. What are the dimensions of such a rectangle with the greatest possible area? Width = Height =

Answer 1

Answer:

Step-by-step explanation:

Given that a rectangle is inscribed with its base on the x-axis and its upper corners on the parabola

$y=2-x^2$

the parabola is open down with vertex at (0,2)

We can find that the rectangle also will be symmetrical about y axis.

Let the vertices on x axis by (p,0) and (-p,0)

Then other two vertices would be (p,2-p^2) (-p,2-p^2) because the vertices lie on the parabola and satisfy the parabola equation

Now width = $2-p^2$

Area = l*w = $2(2p-p^3)$

Use derivative test

I derivative = $2(2-3p^2)$

II derivative = $-12p$

Equate I derivative to 0 and consider positive value only since we want maximum

p = $\sqrt{\frac{2}{3} }$

Thus width= $\sqrt{\frac{2}{3} }$

Length =$2\sqrt{\frac{2}{3} }$

Width = $2-2/3 = 4/3$