Calculus
posted by Anonymous .
3) Consider rectangles located as shown in the first quadrant and inscribed under a decreasing curve, with the lower left hand corner at the origin and the upper right hand corner on the curve
y = sqrt(9  x2)
Find the width, height and area of the largest such rectangle.
let the top right vertex of the rectangle which is on the curve by (x,y)
x is the width of the rectangle, y is the height
then the area = xy
=x√(9x^2)
differentiate using the product rule, set the derivative equal to zero and solve for x.
or....
we could use common sense, notice that your given equation when squared and rearranged is really part of a circle with centre at the origin and radius 3.
So the largest rectangle would be where the x equals the y, namely when x = 3/√2
(which is what you get if you do it by Calculus as I decribed above.
So the area = (3/√2)^2 = 9/2
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