The critical points are and (−1, −1)

Section 13.1 Extreme values 4.

f (x, y) = x⁴+ y⁴− 4xy, f1= 4(x³− y), f2= 4(y³− x) A = f11= 12x², B = f12=−4, C = f22= 12y².

For critical points: x³ = y and y³ = x. Thus x⁹ = x, or x(x⁸− 1) = 0, and x = 0, 1, or−1. The critical points are (0, 0), (1, 1), and (−1, −1). At (0, 0), B²− AC = 16 − 0 > 0, so (0, 0) is a saddle point.At (1, 1) and (−1, 1), B²− AC = 16− 144 < 0, A > 0, so f has local minimum at these points.

5.f (x, y) = ^x_y+_x⁸− y, f1(x, y) = _y¹−_x⁸2 if 8y = x², f2(x, y) =−_y^x2− 1 = 0 if x =

−y².For critical points: 8y = x² = y⁴, so y = 0 or y = 2. f (x, y) is not defined when y = 0, so the only critical point is (−4, 2). At (−4, 2) we have

A = f11= 16 x³ =−1

4, B = f12=−1 y² =−1

4, C = f22=2x y³ =−1.

Thus B²− AC = ₁₆¹ −¹₄ < 0, and (−4, 2) is a local maximum.

7.f (x, y) = xsiny. For critical points we have

f1= siny = 0, f2= xcosy = 0.

Since sinx and cosy cannot vanish at the same point, the only critical points cor- respond to x = 0 and siny = 0. They are (0, nπ), for all integers n. All are saddle points.

11.

f (x, y) = xe^−x3+y3, f₁(x, y) = (1− 3x³)e^−x3+y3, f₂(x, y) = 3xy²e^−x3+y3

A = f11(x, y) = 3x²(3x³− 4)e^−x3+y3, B = f₁₂(x, y) =−3y²(3x³− 1)e^−x3+y3 C = f22(x, y) = 3xy(3y³+ 2)e^−x3+y3

For critical points: 3x³ = 1 and 3xy² = 0. The only critical point is (3⁻¹³, 0).

At that point we have B = C = 0, so the second derivative test is inconclusive.

However, note that f (x, y) = f (x, o)e^y3, and e^y3 has an inflection point at y = 0 .Therefore f (x, y) has neither a maximum nor a minimum value at (3⁻¹³, 0), so has a saddle point there.

14.

f (x, y) = 1

1− x + y + x²+ y² = 1

(x−¹₂)²+ (y +¹₂)²+¹₂ Evidently f has absolute maximum value 2 at (¹₂,−¹₂).Since

f₁(x, y) = 1− 2x

(1− x + y + x²+ y²), f₂(x, y) =− 1 + 2y

(1− x + y + x²+ y²)²(1 2,−1

2)

1

2

is the only critical point of f . 19.

f (x, y) = xye^−x2+y4, f1(x, y) = y(1− 2x²)e^−x2+y4, f2(x, y) = x(1− 4y⁴)e^−x2+y4 For critical points y(1− 2x²) = 0 and x(1− 4y⁴) = 0. The critical points are

(0, 0), (± 1

√2, 1

√2), (± 1

√2,− 1

√2) we have

f (0, 0) = 0, f ( 1

√2, 1

√2) = f (− 1

√2,− 1

√2) = 1

2e⁻³⁴ > 0, f (− 1

√2, 1

√2) = f ( 1

√2,− 1

√2) =−1

2e⁻³⁴ < 0 Since f (x, y)→ 0 as x²+ y² → ∞, the maximum and minimum values of f are

1

2e⁻³⁴, and−¹₂e⁻³⁴ respectively.

20.

f (x, y) = xye^−(x2+y4)

f1(x, y) = y(1− 2x²)e^−(x2+y4), f2(x, y) = x(1− 4y⁴)e^−(x2+y4)

For critical points,x²−y²= 1,and xy = 0. The critical points are (±1, 0).f(±1, 0) =

±¹₂. Since f (x, y)→ 0as x²+ y²→ ∞, the maximum and minimum values of f are

1

2 and−¹₂ respectively.

25.Let(x, y, z) be the coordinates of the corner of the box that is in the first octant of space.Thus x, y, z ≥ 0,and ^x_a²2 + ^y_b²2 + ^z_c²2 = 1. The volume of the box is V = (2x)(2y)(2z) = 8cxy

√

1−^x_a²2 −^y_b²2 for x≥ 0, y ≥ 0 and ^x_a²2 +^y_b2² ≤ 1. For analysis it is easier to deal with V²than with V : V²= 64c²(x²y²−^x_a^4y2²)−^x2_b2^y4.Since V = 0 if x = 0 or y = 0 or ^x_a²₂+^y_b2² = 1, the maximum value of V², and hence of V , will occur at a critical point of V² wherex > 0 and y > 0. For CP:

0 = ∂V²

∂x = 64c²(2xy²−4x³y²

a² −2xy⁴

b² ) = 128c²xy²(1−2x² a² −y²

b²) 0 = ∂V²

∂y = 128c²x²y(1−x² a² −2y²

b² )

Hence we must have ^x_a²2+^2y_b2² = 1 = ^2x_a2²+^y_b2²So that ^x_a²2 = ^y_b2² =¹₃,andx = √^a 3, y =

√b

3. The largest box has volume^8abc

3√

3 cubic units.

27.Differeniate the given equation e^(2zx−x2)− 3e^(2zy+y2) = 2 with respect to x and y, regarding z as a function of x and y:

e^(2zx−x2)(2x∂z

∂x + 2z− 2x) − 3e^(2yz+y2)(2y∂z

∂x) = 0 e^(2zx−x2)(2x∂z

∂y)− 3e^(2zy+y2)(2y∂z

∂y + 2z + 2y) = 0

3

For a critical point we have ^∂z_∂x = 0 and ^∂z_∂y = 0 and it follows form the equations above that z = x and z =−y. Substituting these into the given equation, we get

e^z2− 3e^−z2 = 2, (e^z2)²− 2e^z2− 3 = 0, (e^z2− 3)(e^z2+ 1) = 0.

28.we will use the second derivative test to classify the two critical points calcu- lated in Exercise 25. To calculate the second partials

A = ∂²z

∂x², B = ∂²z

∂x∂y, C = ∂²z

∂y² we obtain

e^2zx−x2[(2x∂z

∂x+ 2z− 2x)²+ 4∂z

∂x+ 2x∂²z

∂x²− 2] − 3e^2yz+y2[(2y∂z

∂x)²+ 2y∂²z

∂x²] = 0.

At a critical point, ^∂z_∂x = 0, z = x, z =−y, andz²= ln3, so 3(2x∂²z

∂x²− 2) − 2 3(2y∂²z

∂x²) = 0 A = ^∂_∂x^2z2 =_6x−2y⁶ .

On the other hand , e^2zx−x2[(2x∂z

∂y)²+ 2x∂²z

∂y²]− 3e^2zy+y2[(2y∂z

∂y + 2z + 2y)²+ 4∂z

∂y + 2y∂²z

∂y² + 2] = 0.

(∂²g

∂y²), 3(2x∂²z

∂y²)−2 3(2y∂²z

∂y² + 2) = 0 C = ∂²z

∂y² = 2 6x− 2y Finally,

e^2zx−x2[(2x∂z

∂x+2z−2x)(2x∂z

∂y)+2x ∂²z

∂x∂y+2∂z

∂y]−3e^2zy+y2[(2y∂z

∂y+2z+2y)(2y∂z

∂x)+2∂z

∂x+2y ∂²z

∂x∂y] = 0.

,and evaluating at a critical point,(6x− 2y)_∂x∂y^∂2z = 0, so that B =_∂x∂y^∂2z = 0. At the critical point(√

ln3,−√

ln3) we have A = _8ln3⁶ , B = 0, C = _8ln3² , so B²− AC < 0, and f has a local minimum at that critical point. At the critical point(−√

ln3,√ ln3) we have A =−_8ln3⁶ , B = 0, C =−_8ln3² ,so B²−AC < 0, and f has a local maximum at that critical point.