Maths help!!!! please urgent
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Maths help!!!! please urgent

[From: ] [author: ] [Date: 12-11-06] [Hit: ]
......
Hi
i don't know how to solve this question please help

A point P lies on the graph of the function y = x2, 0
please help

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The point (x, x^2) determines the width that the height of the rectangle.

width = 2 - x
height = x^2

so the are of the rectangle is
A = width * height = 2x^2 - x^3

A' = 4x - 3x^2

Setting A' = 0 and solving for x, we get

4x - 3x^2 = 0
x(4 - 3x) = 0
x = 0 or x = 4/3

we reject x = 0.
So x = 4/3

The largest possible area of the rectangle is

2(4/3)^2 - (4/3)^3 = 32/27
1
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