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Find the dimensions that give the largest area for a rectangle with its base on the x-axis and its other 2 vertices are above the x-axis, lying on the parabola y=8-x^2 |

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Y = 8 - x² This is symmetrical about the y-axis. So any rectangle fitting inside it has a vertex lying on (x,0). This rectangle has a base length of 2x and a height of y = 8 - x² The area of this rectangle is A = (2x)(8 - x²) = 16x - 2x³. The area is optimized when A' = 0. Since A' = 16 - 6x² we can equate and solve for x 16 = 6x² √(8/3) = x The rectangle has a base of 2√(8/3) and a height of y = 16/3. The area is then (32/3)√(8/3) ≈ 17.418685399 |

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Y = 8 - x² This is symmetrical about the y-axis. So any rectangle fitting inside it has a vertex lying on (x,0). This |

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