LAST NAME : FIRST NAME : QUIZ 4, Version A : MATH 251, Section 506 last name : . . . . . . . . first name : . . . . . . . . GRADE : . . . . . . . . ”An Aggie does not lie, cheat or steal, or tolerate those who do” signature : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Write up your result, detail your calculations if necessary and BOX your final answer Z 1Z x 1. [25pts] Compute the integral 0 sin(x2 )dy dx. 0 2. Let f (x, y) = 4x2 + y 2 − 4x + 2y (a) [25pts] Find the critical points of this function. (b) [50pts] Classify these points ( say if these are maximum, minimum points or saddle points by justifying). 1. We integer the continuous function f (x, y) = sin(x2 ) on {(x, y) ∈ R2 | 0 ≤ x ≤ 1, 0 ≤ y ≤ x}. Z 0 1 Z ( x Z 2 sin(x )dy) dx = 0 1 x y sin(x2 ) 0 dx, 0 Z = 0 = 1 cos(x2 ) x sin(x )dx = − 2 2 1 , 0 1 − cos 1 . 2 2. f is a polynomial function. It admits second partial derivatives at any points in R2 . (a) Determine the critical points of f i.e the sets of points (x, y) such that fx (x, y) = 8x − 4 = 0 = fy (x, y) = 2y + 2. Solve the following system 8x − 4 = 0 ⇔ 2y + 2 = 0 ( 1 2 y = −1, x = 1 We obtain the only one critical point : ( , −1) . 2 (b) Now, classify this point. We have fxx (x, y) = 8, fyy (x, y) = 2, fxy (x, y) = 0 = fyx (x, y) (by the Clairaut’s Theorem since the function is polynomial). 2 = 16 > 0 and fxx ( 21 , −1) = 8 > 0 so ( 12 , −1) is So, fxx ( 12 , −1)fyy ( 21 , −1) − fxy ( 12 , −1) a local minimum point.