Section 14.7(Solutions) 4. f

Section 14.7(Solutions)

4. f_x = 3 − 3x², f_y = −4y + 4y³, f_xx = −6x, f_yy = −4 + 12y², f_xy = 0.

fx = 0, fy = 0 ⇒ critical points are (1, 0), (1, 1), (1, −1), (−1, 0), (−1, 1), (−1, −1).

By Second Derivative Test, f (x, y) has local maximum at (1, 0), local minima at (−1, 1), (−1, −1), and saddle points at (1, 1), (1, −1), (−1, 0).

18. f_x = cos x sin y, f_y = sin x cos y, f_xx = f_yy = − sin x sin y, f_xy = cos x cos y.

f_x = 0, f_y = 0 ⇒ critical points are (0, 0), (^π₂,^π₂), (^π₂, −^π₂), (−^π₂,^π₂), (−^π₂, −^π₂).

By Second Derivative Test, f (x, y) has local maxima f (^π₂,^π₂) = f (−^π₂, −^π₂) = 1, local minima f (^π₂, −^π₂) = f (−^π₂,^π₂) = −1 and saddle point at (0, 0).

20. f_x = e^−x2−y2(2xy − 2x³y), f_y = e^−x2−y2(x²− 2x²y²), f_xx = e^−x2−y2(2y − 10x²y + 4x⁴y), f_yy = e^−x2−y2(−6x²y + 4x²y³), f_xy = e^−x2−y2(2x − 2x³− 4xy²+ 4x³y²).

f_x = 0, f_y = 0 ⇒ critical points are (1, ±^√1

2), (−1, ±^√1

2), (0, y), y ∈ R.

Since f (x, y) → 0 as x² + y² → ∞, by Second Derivative Test, f (x, y) has maximum value at (±1, ^√1

2) and minimum value at (±1, −^√1

2) (f (±1,^√1

2) = ^√1

2e^−3/2 > f (0, y) = 0 > f (±1, −^√1

2) =

−^√1

2e^−3/2). Since f_xy(0, y) = f_yy(0, y) = 0 for all y ∈ R, the infinitely many other critical points (0, y), y ∈ R, satisfy D = 0.

Moreover, just by definition, we know that f (x, y) has local maxima at (0, y) for y < 0, local minima at (0, y) for y > 0, and saddle point at (0, 0).

36. fx = 3x² − 3, fy = −3y² + 12. fx = 0, fy = 0 ⇒ critical points are (±1, 2). f (−1, 2) = 18, f (1, 2) = 14.

Let A = (2, 3), B = (−2, 3), C = (−2, −2), D = (2, 2).

On AB = {(x, 3)| − 2 ≤ x ≤ 2}, f (x, 3) = x³ − 3x + 9 has maxima f (−1, 3) = f (2, 3) = 11 and minima f (−2, 3) = f (1, 3) = 7.

On BC = {(−2, y)| − 2 ≤ y ≤ 3}, f (−2, y) = −y³ + 12y − 2 has maximum f (−2, 2) = 14 and

1

minimum f (−2, −2) = −18.

On CD = {(t, t)| − 2 ≤ t ≤ 2}, f (t, t) = 9t has maximum f (2, 2) = 18 and minimum f (−2, −2) =

−18.

On DA = {(2, y)|2 ≤ y ≤ 3}, f (2, y) = −y³ + 12y + 2 has maximum f (2, 2) = 18 and minimum f (2, 3) = 11.

Hence f (x, y) has maxima f (−1, 2) = f (2, 2) = 18 and minimum f (−2, −2) = −18.

42. We want to minimize x² + y² + z² = x² + 9 + xz + z² with y² = 9 + xz ≥ 0. Let f (x, z) = x²+ xz + z²+ 9, then f_x = 2x + z and f_z = x + 2z. So the critical point of f (x, z) is (x, z) = (0, 0).

On the boundary {(x, z)|xz = 9}, f (x, z) = x²+ z² ≥ 2|xz| = 18. By Second Derivative Test, (0, 0) is the only local minimum. But f (x, z) is bounded below, hence it is minimum. Hence the minimum distance is pf(0, 0) = 3 and the points closest to the origin are (0, ±3, 0).

56. Suppose the equation of the plane is ax + by + cz = 1, a, b, c > 0. Since the plane passes through (1, 2, 3), we have a + 2b + 3c = 1. We want to minimize the volume _6abc¹ , so we have to maximize 6abc = 2ab(1 − a − 2b) = 2ab − 2a²b − 4ab² with a, b, 1 − a − 2b > 0. Let f (a, b) = 2ab − 2a²b − 4ab², then f_a = 2b(1 − 2a − 2b) and f_b = 2a(1 − a − 4b). f_a = 0, f_b = 0 ⇒ critical point of f (a, b) is (a, b) = (¹₃,¹₆). Since f (¹₃,¹₆) = ₂₇¹ is local maximum(by Second Derivative Test) and the value of f tends to 0 as (a, b) tends to the boundary of {(a, b)|a > 0, b > 0, 1 − a − 2b > 0}, so f (a, b) has maximum ₂₇¹ . Hence the minimum volume is _{f (}1¹

3,¹₆) = 27.

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