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

Take x > 0. Then for 0 < x < √2, a rectangle circumscribed by the x-axis and the parabola y = 2 - x² has a length of 2x and a height of 2 - x², so its area is

A(x) = 2x (2 - x²)

Differentiate A with respect to x :

A'(x) = 2 (2 - x²) + 2x (-2x)

A'(x) = 4 - 6x²

Find the critical points:

4 - 6x² = 0

6x² = 4

x² = 4/6 = 2/3

x = √(2/3)

Check the sign of the second derivative at this point to confirm that it's a maximum:

A''(x) = -12x

The second derivative is negative for all x > 0, so x = √(2/3) is indeed a maximum.

So the rectangle with maximum area has dimensions

width/length = 2 √(2/3)

height = 2 - (√(2/3))² = 4/3

the answer is the third / last answer

step-by-step explanation:

48 / 6 = x

in this case, x will equal the amount of money in each card!

x therefore will be 8.

trent had $8 in each birthday card.