By James Brown, Ruel Churchill

Advanced Variables and purposes, 8E

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**Example text**

The points in T must satisfy the inequalities 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ 4x − 4y. In particular, the triangle B is B = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x}, and we get T = {(x, y, z) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ 4x − 4y}. 4 x Figure 7: The domain B The plane z = constant ∈ [0, 4] cuts the tetrahedron in a triangle which is similar to B of the similarity factor 1 − z4 for z ∈ [0, 4]), thus B(z) = (x, y) z 0≤x≤1− , 0≤y ≤x , 4 and accordingly, T = (x, y, z) z 0 ≤ z ≤ 4, 0 ≤ x ≤ 1 − , 0 ≤ y ≤ x . 4 5) The obvious planes are y = 0, z = 0, [points (1, 0, 0), (−1, 0, 0), (0, 0, 4)], [points (1, 0, 0), (0, 2, 0), (−1, 0, 0)].

Since Ga (x) → −∞ for thus G√ a is increasing in ] − x → ± a2 − 1, the range of Ga is given by 1+ ] − ∞, Ga (0)] = −∞, ln(a − 1) − a −1 . 2 3) It follows from (2) when y = a that dGa (0) 1 1 = − , da a−1 2 Please click the advert Student Discounts + Student Events + Money Saving Advice = Happy Days! com 57 Calculus 2c-1 Examinations of functions hence Ga (0) is increasing for a ∈ ]1, 3[ and decreasing for a ∈ ]3, +∞[. Since Ga (0) → −∞ for a → 1+ and for a → +∞, we see that G3 (0) = f (0, 3) = ln 2 − 5 2 is a global maximum.

In fact, there are no stationary points, and if x 2 + y 2 → +∞, then at least one of the two expressions |x − y| or |x + y| will tend towards +∞. 4 Prove that the function g(u) = u2 exp −u2 , u ∈ [0, +∞[, has a maximum for u = 1, such that g(u) < g(1) for u = 1. Then ﬁnd the maximum and minimum of the function f (x, y) = x2 exp −x2 − y 2 , (x, y) ∈ R2 . A Maximum and minimum. D Diﬀerentiate g(u). If possible, apply a substitution. Either write f (x, y) = g(x) exp −y 2 , or apply polar coordinates.