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A rectangle has its base on the x-axis and its two upper vertices on…
Question
A rectangle has its base on the x-axis and its two upper vertices on the parabola y = 12 - x^2 (with y >= 0). Find the maximum possible area of such a rectangle.
✓ Verified answer: 32checked by our engine — not a guess
Step-by-step solution
By symmetry the upper vertices are at (±x, 12 - x^2), so the rectangle has width 2x and height 12 - x^2, giving area A(x) = 2x(12 - x^2) = 24x - 2x^3 for 0 < x < sqrt(12).
Then A'(x) = 24 - 6x^2 = 0 gives x^2 = 4, x = 2.
A''(x) = -12x < 0, a maximum.
A(2) = 2·2·(12 - 4) = 4·8 = 32.
Final answer32
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