The largest rectangle inscribed under the parabola y = 4 - x^2, base on the x-axis, has area

\tfrac{32\sqrt 3}{9} A(x) = 2x(4 - x^2). A' = 8 - 6x^2 = 0 \Rightarrow x = 2/\sqrt 3. A = 2(2/\sqrt 3)(4 - 4/3) = (4/\sqrt 3)(8/3) = 32/(3\sqrt 3) = 32\sqrt 3/9.

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Acemy

Optimization

AP Calculus AB· difficulty 3/5

y = 4 - x^2

A
$8$
B
$\tfrac{8\sqrt 3}{3}$
C
$\tfrac{16\sqrt 3}{9}$
D
$\tfrac{32\sqrt 3}{9}$
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Explanation

A(x) = 2x(4 - x^2)

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