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=
i figured out height was 4/3 cause i thought width was
sqrt(2/3)
and plugged it in y=2-x^2

Question

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=
i figured out height was 4/3 cause i thought width was 
sqrt(2/3)
 and plugged it in y=2-x^2

amistre64 · Answer

Area = xy

y = 2-x^2

Area = x(2-x^2)

Area = 2x - x^3

derive and conquer :)

amistre64 · Answer

A' = 2 - 3x^2

2 -3x^2 = 0

2 = 3x^2

2/3 = x^2

+-sqrt(2/3) = x

amistre64 · Answer

x = sqrt(6)/3

y=2-(sqrt(6)/3)^2

y = 2 - (6/9)

y = 18/9 - 6/9

y = 12/9 = 4/3

Max area:

4sqrt(6)
------
    9

i think :)