Question

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=4−x2. What are the dimensions of such a rectangle with the greatest possible area?

Answer 1

2sqrt3/3 by 8/3

Answer 2

l=2*(4/3)^0.5
h=4-(4/3)^2=8/3

Answer 3

myininaya can u come help me when u r finished?

Answer 4

base is a-(-a)=a+a=2a
height is y=4-a^2

Area=base*height=2a(4-a^2)
Area=8a-2a^3

A'=8-6a^2
set A'=0
8-6a^2=0
solve for a
8=6a^2
divide 6 on both sides
8/6=a^2

4/3=a^2

so

\[a=\pm \sqrt{\frac{4}{3}}=\pm \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}=\pm \frac{2 \sqrt{3}}{3}\]

so we have
\[a=\frac{2 \sqrt{3}}{3}\]