Explanation:
you mean A = area of the rectangle ?
and p = the perimeter of the rectangle ?
the area of the rectangle is
length×width = 2y × 2x = 4xy
the perimeter is
2×length + 2×width = 2×2y + 2×2x = 4x + 4y
based on the primary circle function
x² + y² = 4
we get
y² = 4 - x²
y = sqrt(4 - x²)
1)
f(x) = 4xy = 4x×sqrt(4 - x²) = g(x)×h(x)
g(x) = 4x
h(x) = h(i(x)) = sqrt(4 - x²)
i(x) = 4 - x²
the zeros of the first derivation gives us the extreme point(s).
f'(x) = g'(x)×h(x) + g(x)×h'(x)
g'(x) = 4
(h(i(x))' = h'(i(x))×i'(x)
h'(x) = 1/(2×sqrt(x))
i'(x) = -2x
so,
f'(x) = 4×sqrt(4 - x²) + 4x×-2x/(2×sqrt(4 - x²)) =
= 4×sqrt(4 - x²) - 4x²/sqrt(4 - x²)
f'(x) = 0
4×sqrt(4 - x²) - 4x²/sqrt(4 - x²) = 0, x² < 4, -2 < x < 2
sqrt(4 - x²) - x²/sqrt(4 - x²) = 0
sqrt(4 - x²) = x²/sqrt(4 - x²)
let's multiply by sqrt(4 - x²)
4 - x² = x²
4 = 2x²
x² = 2
x = sqrt(2)
therefore, the largest A (area of the rectangle) is for
x = sqrt(2), which also makes y = sqrt(2), which makes the rectangle a square.
2)
f(x) = 4x + 4×sqrt(4 - x²)
f'(x) = 4 + 4×-2x/(2×sqrt(4 - x²)) = 0, -2 < x < 2
4 - 4x/sqrt(4 - x²) = 0
1 - x/sqrt(4 - x²) = 0
1 = x/sqrt(4 - x²)
multiply by sqrt(4 - x²)
sqrt(4 - x²) = x
square both sides
4 - x² = x²
4 = 2x²
x = sqrt(2)
so, also the perimeter of the rectangle is at maximum with
x = sqrt(2) and the rectangle is again a square.