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Unformatted text preview: MATH 31A (Butler) Practice for Midterm IIa (Solutions) 1. (a) Verify that (2 , 1) is a critical point for the curve y 3 x 3 yx 2 = x 2 15 x + 16. (You need to verify two things: (i) it is on the curve and (ii) it is a critical point.) To verify it is a point on the curve we plug x = 2 and y = 1 into both sides and we get 1 3 2 3 1 2 2 = 10 = 2 2 15 2 + 16 . So it is a point on the curve. To see that it is a critical point we first compute the derivative using implicit differentiation, i.e., 3 y 2 y x + y 3 3 y x 2 6 xy = 2 x 15 or (3 y 2 x 3 x 2 ) y = 2 x +6 xy 15 y 3 . Plugging x = 2 and y = 1 into this equation we have (3 1 2 2 3 2 2 ) y = 2 2 + 6 2 1 15 1 3 or 6 y = 0 , and so we can conclude at x = 2 and y = 1 that y = 0, i.e., it is a critical point. (b) Use the second derivative test to determine if the point (2 , 1) is a maximum or a minimum. To use the second derivative test we need the second derivative. So starting with (3 y 2 x 3 x 2 ) y = 2 x + 6 xy 15 y 3 (from above) we take the derivative of both sides to get the following (6 yy x + 3 y 2 6 x ) y + (3 y 2 x 3 x 2 ) y 00 = 2 + 6 xy + 6 y 3 y 2 y . Now we can substitute in x = 2, y = 1 and y = 0 so that we have (6 1 2 + 3 1 2 6 2) 0 + (3 1 2 2 3 2 2 ) y 00 = 2 + 6 2 0 + 6 1 3 1 2 , which simplifies to 6 y 00 = 8 so that y 00 = 4 / 3 < 0. So by the second derivative test the curve is concave down at (2 , 1) and so (2 , 1) is a maximum. 2. You have recently started a new job as quality control officer in the new widget factory, company slogan: You fidget, we widget. A new machine has been installedfactory, company slogan: You fidget, we widget....
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 Winter '10
 BUTLER
 Critical Point

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