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l=2x

h=6-x2

A=lh=2x(6-x2)=12x-2x3

A will have a maximum when it's derivative is 0.

dA/dx = 12-6x2 = 0

12=6x2

x=+/- sqrt(2)

Since the length of the rectangle has to be positive, x = sqrt(2)

A = 2*sqrt(2)*(6-2) = 8*sqrt(2) ~= 11.314

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Q: What is the answer A rectangle has its base on the x -axis and its upper vertices on the parabola y equals 6-X squared Find the largest area the rectangle can have?
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