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A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 11 − x 2 . What are the dimensions of such a rectangle with the greatest possible area?

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Answer:
(22)/(3),
2\dot \sqrt{(11)/(3)}

Explanation:

Given

rectangle with its base on x-axis

and other two corners at parabola

and parabola is downward facing symmetric about y-axis

let y be the y co-ordinate of the corner thus x co-ordinate is given by


x=\pm √(11-y)

Thus lengths of rectangle is
2√(11-y) & y

Area
=y* 2√(11-y)

differentiating w.r.t to y for maximum area


\frac{\mathrm{d} A}{\mathrm{d} y}=2* √(11-y)-(y)/(2\dot √(11-y))=0

we get y=
(22)/(3)

and
x=\pm \sqrt{(11)/(3)}

A_{max}=16.21 units

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