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Advanced MathQ&A Library1. LOCAL VERSUS ABSOLUTE EXTREMA. You might expect from single-variable calculus that if a function has only one critical point, and that critical point is a local minimum (say), then that critical point is the global/absolute minimum. This fails spectacularly in higher dimensions (and there's a famous example of a mistake in a mathematical physics paper because this fact was not properly appreciated.) You will compute a simple example in this problem. Let f(x, y) = e* + y³ – 3ye". (a) Find all critical points for this function; in so doing you will see there is only one. (b) Verify this critical point is a local minimum. (c) Show this is not the absolute minimum by finding values of f(r, y) that are lower than the value at this critical point. We suggest looking at values f(0, y) for suitably chosen y. 2. The distance from (x, y, z) to the origin is Vr? + y? + 22. We want to minimize it, which is equivalent to minimize f(x, y, z) = x² + y? + z?. (a) Amongst all the points on the plane r – 2y + 3z = 6, there is a unique point that is closest to the origin. Find this point using Lagrange multipliers, and find the distance of this point to the origin. (b) Earlier we learned a formula for the distance from a point to a plane. Apply this formula to verify your answer. 3. Let f(r, y) = a² + y?. (a) Find the point that satisfies the Lagrange multiplier condition for f(x, y) subject to xy = 9, with a> 0 and y > 0 (first quadrant). (b) Draw the constraint curve ry = 9 in the first quadrant, and label the point you found in the previous part. Draw the contour of f(r, y) through this point. (c) Draw more contours of f(r, y). Use them to show the point you found is the absolute minimum of f(x, y) subject to ry = 9 (first quadrant), and that there is no maximum. 1Start your trial now! First week only $4.99!*arrow_forward*

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