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Let y=2-x^2. Choose a point (x,y) in the first quadrant that is on this curve. Then form a rectangle in the first quadrant using the points (x,y) and (0,0) as the ends of a diagonal for the rectangle. Find the values of x and y which make the area of the rectangle the largest.
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area of rectangle is xy A=x(2-x^2) =2x-x^3 diff A w.r.t x to get extremum as x=\[\sqrt{2/3}\] ==>y=4/3
maximize f(x,y) = xy for constraint y = 2 - x^2 might be this http://www.wolframalpha.com/input/?i=maximize+x%282-x^2%29 http://www.wolframalpha.com/input/?i=2+-+2%2F3 says it lies in first quadrant
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