Calculus
posted by Marissa .
A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y= 12  x^2. What are the dimensions of such a rectangle with the greatest possible area?

The vertices of the rectangle are points:
(x,0),(x,0),(x,12x^2),(x,12x^2).
The area A(x)=2x(12x^2)
A(x)=24x2x^3
A'(x)=246x^2=0 => x=+2
max A(x)=A(2)=4816=32
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