EliteGenius From: US Posts: 13 Votes: 39 |
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=8−x2. What are the dimensions of such a rectangle with the greatest possible area? Width = ? Height = ? |

Anonymous From: - Posts: - Votes: - |
The corners of the base are located at x and -x now height will be equal to y area = y*(x-(-x)) = 2xy = 2x(8-x^2) for area to be maximum dArea/dx shoud be zero = 16-6x^2 = 0 x= 1.6329 y = 5.333 area = 2xy = 17.41 |

LondonSam9 From: US Posts: 40 Votes: 40 |
Let the width of rectangle = 2x, now height becomes the value of y at x = 8-x^2 area = 2x*(8-x^2) = 16x - 2x^3 differentiate with respect to x and equate to 0, we get x = + or - 1.633 Therefore required width = 2*1.633 = 3.265, height = 5.33 |

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