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oh sorry the parabola is y=5-x^2
I was thinking find the 0 for the parabola, and because it touches the 2 upper part I get sqrt(5) and -sqrt(5) being the length for y and I can ind the length in x because those 2 y are the distance between 2 cornor that is the length?
Since the rectangular inscribed in the parabola:
it's length is symmetry x+ x
=> A = 2x y
= 2x ( 5 - x²)
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oh. inscribed in, so i had it wrong... so you don't have any constraint in this problem...?
I think the range of parabola is the constraint, just get the concept first!
Well I usually go about these problem by finding my constraint then the equation I need to solve, but I couldn't do so.
But when you say A=x(5-x^2) and so I take the derivative of both side and I should get
x(5-2x) + (5-x^2) = A'(x)
-3x^2+5x+5 and then use quadratic to find my x ?