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 =

Step-by-step explanation:

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

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 =

Area = l*w =

Use derivative test

I derivative =

II derivative =

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

p =

Thus width=

Length =

Width =

angle addition postulate

2x=56

step-by-step explanation:

351°f

step-by-step explanation: