Question Two: a. Define the following: i. Local/Relative Maximum ii. Local/Relative Minimum iii. Critical Point. b. Find and classify the relative extrema for the function F(x,y)=-x2+4y-2y3+1. c. Find the absolute maximum/minimum of f(x,y)=4x+6y-x2-y2 on the boundary B = [(x,y)|0<=x<=4,0<=y<=5].

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Asked on June 7, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-06-07T06:51:00.128Z

Question Two a) Define the following: i. Local/Relative Maximum: A function f(x,y) has a local maximum at a point (x_0, y_0) if f(x,y) f(x_0, y_0) for all points (x,y) in an open disk centered at (x_0, y_0). ii. Local/Relative Minimum: A function f(x,y) has a local minimum at a point (x_0, y_0) if f(x,y) f(x_0, y_0) for all points (x,y) in an open disk centered at (x_0, y_0). iii. Critical Point: A point (x_0, y_0) is a critical point of a function f(x,y) if its first partial derivatives f_x(x_0, y_0) and f_y(x_0, y_0) are both equal to zero, or if one or both of the partial derivatives do not exist at (x_0, y_0). b) Find and classify the relative extrema for the function F(x,y) = -x^2 + 4xy - 2y^2 + 1. Step 1: Find the first partial derivatives of F(x,y). F_x = ()/( x)(-x^2 + 4xy - 2y^2 + 1) = -2x + 4y F_y = ()/( y)(-x^2 + 4xy - 2y^2 + 1) = 4x - 4y Step 2: Set the partial derivatives to zero to find critical points. -2x + 4y = 0 x = 2y (1) 4x - 4y = 0 x = y (2) Substitute (2) into (1): y = 2y y = 0 Then x = 0. The only critical point is (0,0). Step 3: Use the Second Derivative Test to classify the critical point. First, find the second partial derivatives: F_xx = ()/( x)(-2x + 4y) = -2 F_yy = ()/( y)(4x - 4y) = -4 F_xy = ()/( y)(-2x + 4y) = 4 Calculate the discriminant D = F_xxF_yy - (F_xy)^2: D = (-2)(-4) - (4)^2 = 8 - 16 = -8 Since D < 0, the critical point (0,0) is a saddle point. Therefore, there are no relative maxima or minima for the function F(x,y). The relative extrema are: None c) Find the absolute maximum/minimum of f(x,y) = 4x + 6y - x^2 - y^2 on the boundary B = \(x,y) | 0 x 4, 0 y 5\. Step 1: Find critical points inside the region. f_x = ()/( x)(4x + 6y - x^2 - y^2) = 4 - 2x f_y = ()/( y)(4x + 6y - x^2 - y^2) = 6 - 2y Set f_x = 0 4 - 2x = 0 x = 2. Set f_y = 0 6 - 2y = 0 y = 3. The critical point is (2,3). This point lies within the given region 0 x 4, 0 y 5. Step 2: Evaluate f(x,y) at the critical point. f(2,3) = 4(2) + 6(3) - (2)^2 - (3)^2 = 8 + 18 - 4 - 9 = 26 - 13 = 13 Step 3: Evaluate f(x,y) along the boundaries of the region. The region is a rectangle with corners (0,0), (4,0), (0,5), (4,5). Boundary 1: x=0, 0 y 5 f(0,y) = 6y - y^2. Let g(y) = 6y - y^2. g'(y) = 6 - 2y = 0 y = 3. Points to check: (0,3), and the endpoints (0,0), (0,5). f(0,3) = 6(3) - (3)^2 = 18 - 9 = 9. f(0,0) = 0. f(0,5) = 6(5) - (5)^2 = 30 - 25 = 5. Boundary 2: x=4, 0 y 5 f(4,y) = 4(4) + 6y - (4)^2 - y^2 = 16 + 6y - 16 - y^2 = 6y - y^2. This is the same as Boundary 1. Critical point at y=3. Points to check: (4,3), and the endpoints (4,0), (4,5). f(4,3) = 6(3) - (3)^2 = 18 - 9 = 9. f(4,0) = 0. f(4,5) = 6(5) - (5)^2 = 30 - 25 = 5. Boundary 3: y=0, 0 x 4 f(x,0) = 4x - x^2. Let h(x) = 4x - x^2. h'(x) = 4 - 2x = 0 x = 2. Points to check: (2,0), and the endpoints (0,0), (4,0). f(2,0) = 4(2) - (2)^2 = 8 - 4 = 4. f(0,0) = 0 (already found). f(4,0) = 0 (already found). Boundary 4: y=5, 0 x 4 f(x,5) = 4x + 6(5) - x^2 - (5)^2 = 4x + 30 - x^2 - 25 = 4x - x^2 + 5. Let k(x) = 4x - x^2 + 5. k'(x) = 4 - 2x = 0 x = 2. Points to check: (2,5), and the endpoints (0,5), (4,5). f(2,5) = 4(2) - (2)^2 + 5 = 8 - 4 + 5 = 9. f(0,5) = 5 (already found). f(4,5) = 5 (already found). Step 4: Compare all candidate values. The values of f(x,y) at the critical point and boundary points are: f(2,3) = 13 f(0,3) = 9 f(0,0) = 0 f(0,5) = 5 f(4,3) = 9 f(4,0) = 0 f(4,5) = 5 f(2,0) = 4 f(2,5) = 9 The maximum value is 13. The minimum value is 0. The absolute maximum is 13 and the absolute minimum is 0. What's next?

Accepted Answer · 2026-06-07T06:51:00.128Z

Question Two a) Define the following: i. Local/Relative Maximum: A function f(x,y) has a local maximum at a point (x_0, y_0) if f(x,y) f(x_0, y_0) for all points (x,y) in an open disk centered at (x_0, y_0). ii. Local/Relative Minimum: A function f(x,y) has a local minimum at a point (x_0, y_0) if f(x,y) f(x_0, y_0) for all points (x,y) in an open disk centered at (x_0, y_0). iii. Critical Point: A point (x_0, y_0) is a critical point of a function f(x,y) if its first partial derivatives f_x(x_0, y_0) and f_y(x_0, y_0) are both equal to zero, or if one or both of the partial derivatives do not exist at (x_0, y_0). b) Find and classify the relative extrema for the function F(x,y) = -x^2 + 4xy - 2y^2 + 1. Step 1: Find the first partial derivatives of F(x,y). F_x = ()/( x)(-x^2 + 4xy - 2y^2 + 1) = -2x + 4y F_y = ()/( y)(-x^2 + 4xy - 2y^2 + 1) = 4x - 4y Step 2: Set the partial derivatives to zero to find critical points. -2x + 4y = 0 x = 2y (1) 4x - 4y = 0 x = y (2) Substitute (2) into (1): y = 2y y = 0 Then x = 0. The only critical point is (0,0). Step 3: Use the Second Derivative Test to classify the critical point. First, find the second partial derivatives: F_xx = ()/( x)(-2x + 4y) = -2 F_yy = ()/( y)(4x - 4y) = -4 F_xy = ()/( y)(-2x + 4y) = 4 Calculate the discriminant D = F_xxF_yy - (F_xy)^2: D = (-2)(-4) - (4)^2 = 8 - 16 = -8 Since D < 0, the critical point (0,0) is a saddle point. Therefore, there are no relative maxima or minima for the function F(x,y). The relative extrema are: None c) Find the absolute maximum/minimum of f(x,y) = 4x + 6y - x^2 - y^2 on the boundary B = \(x,y) | 0 x 4, 0 y 5\. Step 1: Find critical points inside the region. f_x = ()/( x)(4x + 6y - x^2 - y^2) = 4 - 2x f_y = ()/( y)(4x + 6y - x^2 - y^2) = 6 - 2y Set f_x = 0 4 - 2x = 0 x = 2. Set f_y = 0 6 - 2y = 0 y = 3. The critical point is (2,3). This point lies within the given region 0 x 4, 0 y 5. Step 2: Evaluate f(x,y) at the critical point. f(2,3) = 4(2) + 6(3) - (2)^2 - (3)^2 = 8 + 18 - 4 - 9 = 26 - 13 = 13 Step 3: Evaluate f(x,y) along the boundaries of the region. The region is a rectangle with corners (0,0), (4,0), (0,5), (4,5). Boundary 1: x=0, 0 y 5 f(0,y) = 6y - y^2. Let g(y) = 6y - y^2. g'(y) = 6 - 2y = 0 y = 3. Points to check: (0,3), and the endpoints (0,0), (0,5). f(0,3) = 6(3) - (3)^2 = 18 - 9 = 9. f(0,0) = 0. f(0,5) = 6(5) - (5)^2 = 30 - 25 = 5. Boundary 2: x=4, 0 y 5 f(4,y) = 4(4) + 6y - (4)^2 - y^2 = 16 + 6y - 16 - y^2 = 6y - y^2. This is the same as Boundary 1. Critical point at y=3. Points to check: (4,3), and the endpoints (4,0), (4,5). f(4,3) = 6(3) - (3)^2 = 18 - 9 = 9. f(4,0) = 0. f(4,5) = 6(5) - (5)^2 = 30 - 25 = 5. Boundary 3: y=0, 0 x 4 f(x,0) = 4x - x^2. Let h(x) = 4x - x^2. h'(x) = 4 - 2x = 0 x = 2. Points to check: (2,0), and the endpoints (0,0), (4,0). f(2,0) = 4(2) - (2)^2 = 8 - 4 = 4. f(0,0) = 0 (already found). f(4,0) = 0 (already found). Boundary 4: y=5, 0 x 4 f(x,5) = 4x + 6(5) - x^2 - (5)^2 = 4x + 30 - x^2 - 25 = 4x - x^2 + 5. Let k(x) = 4x - x^2 + 5. k'(x) = 4 - 2x = 0 x = 2. Points to check: (2,5), and the endpoints (0,5), (4,5). f(2,5) = 4(2) - (2)^2 + 5 = 8 - 4 + 5 = 9. f(0,5) = 5 (already found). f(4,5) = 5 (already found). Step 4: Compare all candidate values. The values of f(x,y) at the critical point and boundary points are: f(2,3) = 13 f(0,3) = 9 f(0,0) = 0 f(0,5) = 5 f(4,3) = 9 f(4,0) = 0 f(4,5) = 5 f(2,0) = 4 f(2,5) = 9 The maximum value is 13. The minimum value is 0. The absolute maximum is 13 and the absolute minimum is 0. What's next?

Question Two: a. Define the following: i. Local/Relative Maximum ii. Local/Relative Minimum iii. Critical Point. b. Find and classify the relative extrema for the function F(x,y)=-x2+4y-2y3+1. c. Find the absolute maximum/minimum of f(x,y)=4x+6y-x2-y2 on the boundary B = [(x,y)|0<=x<=4,0<=y<=5].

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Question Two: a. Define the following: i. Local/Relative Maximum ii. Local/Relative Minimum iii. Critical Point. b. Find and classify the relative extrema for the function F(x,y)=-x2+4y-2y3+1. c. Find the absolute maximum/minimum of f(x,y)=4x+6y-x2-y2 on the boundary B = [(x,y)|0<=x<=4,0<=y<=5].

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