Step 1:
Find the interval of x.
y = -4x² + 4
0 = -4x² + 4
-4 = -4x²
x² = 1
x = ±1
The interval for x is 0 ≤ x ≤ 1.
Step 2:
Maximize the area of the rectangle.Area = lw
w = 2x
l = y = (-4x² + 4) - 0
A(x) = 2x(-4x² + 4)
A(x) = -8x³ + 8x
A’(x) = -24x² + 8
0 = -24x² + 8
-8 = -24x²
x² = 1/3
x =
The First Derivative Test:
A’(x):
Conclusion: Therefore, the area has an absolute Maximum at x = .
Part "a":
Part "a" Answer: The coordinates of C are 
.
Part b. x’ = 2. Find A’(x) when x = ½.
A(x) = -8x³ + 8x
A’(x) = -24x²x’ + 8x’
A’(1/2) = -24(1/2)²(2)+ 8(2)
A’(1/2) = -24(1/4) (2)+ 16
A’(1/2) = -14 + 16 = 4
Part "b" Answer
: A is changing at the rate of 4 square units per second.