Asked by Jean
Find the rectangle with the maximum area inside the ellipse x^2/16 + y^2/4=1.
Thanks.
Thanks.
Answers
Answered by
Steve
If the base of the rectangle has length 2x, and height y, the area is
a = 2xy = 2x√(1-x^2/16)(4)
= 2x√(16-x^2)
da/dx = 4(8-x^2)/√(16-x^2)
so, max area occurs at x=√8, and the rectangle is 2√8 by √2
a = 2xy = 2x√(1-x^2/16)(4)
= 2x√(16-x^2)
da/dx = 4(8-x^2)/√(16-x^2)
so, max area occurs at x=√8, and the rectangle is 2√8 by √2
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