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exam2f04sol - MATH 215 Fall 2004 Solutions to the Second...

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MATH 215 – Fall 2004 Solutions to the Second Exam Problem 1. (10+5 = 15 points) This problem is about the iterated double integral I = Z 2 0 Z 1 y/ 2 y exp( x 5 ) dx dy (a) Sketch the region of integration and change the order of integration. Solution. The domain of integration is the region above the x axis, between the lines x = 0 and x = 1, and below the parabola y = 2 x 2 . The result of changing the order of integration is: I = Z 1 0 Z 2 x 2 0 y exp( x 5 ) dy dx. / (b) Evaluate I by computing the double integral of your answer to part (a). Solution. Evaluating the integral with respect to y one obtains: Z 2 x 2 0 y exp( x 5 ) dy = exp( x 5 ) Z 2 x 2 0 y exp( x y ) dy = 2 exp( x 5 ) x 4 , so I = 2 Z 1 0 x 4 exp ( x 5 ) dx . If u = x 5 , x 5 dx = du/ 5, and so I = 2 5 Z 1 0 e u du = 2 5 ( e - 1) . / Problem 2. (10+15=25 points) (a) Using the method of Lagrange multipliers, find the area of the largest rectangle with pairs of sides parallel to the coordinate axes that can be inscribed in the ellipse x 2 + 4 y 2 = 1. Also give the coordinates of the corner of the rectangle in the first quadrant. Solution. If ( x, y ) denote the coordinates of a point in the first quadrant, the area of the rectangle with a corner at ( x, y ), sides parallel to the coordinate axes and center at the origin is: f ( x, y ) = (2 x )(2 y ) = 4 xy . The problem is to maximize this function subject to the constraint: x 2 + 4 y 2 = 1, so that the constraining function is g ( x, y ) = x 2 + 4 y 2 - 1. The vector equation f = λ g becomes the system: 4 y = 2 λx 4 x = 8 λy. Substituting into the 1st equation we get: 2 y = 2 λ 2 y . Clearly we are not interested in the solution y = 0, for then the area is zero. We can therefore divide this equation by y , and get λ = ± 1. Since we are looking for a solution with both x, y positive we must have λ = 1. Substituting x = 2 y into the constraint equation one gets: 1 = 4 y 2 + 4 y 2 , and therefore y = 2 4 since y > 0. Since x = 2 y , the final conclusion is: The coordinates of the corner are: ( 2 2 , 2 4 ) ; the maximal area is 1 . / 1
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