This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 Z 1 y/ 2 y exp( x 5 ) dxdy (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 Z 2 x 2 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 y exp( x 5 ) dy = exp( x 5 ) Z 2 x 2 y exp( x y ) dy = 2 exp( x 5 ) x 4 , so I = 2 Z 1 x 4 exp ( x 5 ) dx . If u = x 5 , x 5 dx = du/ 5, and so I = 2 5 Z 1 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 = g becomes the system: 4 y = 2 x 4 x = 8 y. Substituting into the 1st equation we get: 2 y = 2 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 = 1. Since we are looking for a solution with both x, y positive we must have = 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...
View
Full
Document
This note was uploaded on 04/11/2008 for the course MATH 215 taught by Professor Fish during the Fall '08 term at University of Michigan.
 Fall '08
 Fish
 Math

Click to edit the document details