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3. Examine the following 3-dimensional surfaces for maximum, minimum or saddle points: (a) Z(х, у) %3 Зx? — 4ху + 10х + бу?— 44у |

3. Examine the following 3-dimensional surfaces for maximum, minimum or saddle points: (a) Z(х, у) %3 Зx? — 4ху + 10х + бу?— 44у |

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Can you please solve Question 3 a and b? I wish to compare it to my own workings. Also please find a similar question with the awnser underneath to make it easier for you to know what is necessary. 

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3. Examine the following 3-dimensional surfaces for maximum, minimum or saddle points: (a) Z (х, у) %3 Зx2 — 4ху + 10х + 6у2 — 44у Z(x, y) = 2x² – xy +y² – x³ (b) -
Transcribed Image Text:3. Examine the following 3-dimensional surfaces for maximum, minimum or saddle points: (a) Z (х, у) %3 Зx2 — 4ху + 10х + 6у2 — 44у Z(x, y) = 2x² – xy +y² – x³ (b) -
4. Examine the following 3-dimensional surfaces for maximum, minimum or saddle points. (a) Z(x, y) = 1.5x² – 2xy + 5x + 3y? – 22y Z, %3D -2х + бу — 22 %3D0 IOC Solves x = 1, y = 4 Zx = 3; Zyy = 6; Zxy = -2 (Z)(Zyy) > (Zxy)² Satisfies the conditions for a min (b) Z(x, y) = 4x² – 2xy + y² – 2x² Zx = 4x – 2y – 6x? = 0 Zy = -2x + 2y = 0 IOC The second equation gives us x = y. Substitute into the first equation and solve to find x = 0 (y =0) and x = 1/3 (y = 1/3). Zx = 4 – 12x; Zyy = 2; Zgy – 2 Testing the first point on the surface we get a minimum, testing the second point we get a point which is neither max nor min. But we cannot say it's a saddle point (we'd need the second order partials to be opposite sign, in this case Z = 0).
Transcribed Image Text:4. Examine the following 3-dimensional surfaces for maximum, minimum or saddle points. (a) Z(x, y) = 1.5x² – 2xy + 5x + 3y? – 22y Z, %3D -2х + бу — 22 %3D0 IOC Solves x = 1, y = 4 Zx = 3; Zyy = 6; Zxy = -2 (Z)(Zyy) > (Zxy)² Satisfies the conditions for a min (b) Z(x, y) = 4x² – 2xy + y² – 2x² Zx = 4x – 2y – 6x? = 0 Zy = -2x + 2y = 0 IOC The second equation gives us x = y. Substitute into the first equation and solve to find x = 0 (y =0) and x = 1/3 (y = 1/3). Zx = 4 – 12x; Zyy = 2; Zgy – 2 Testing the first point on the surface we get a minimum, testing the second point we get a point which is neither max nor min. But we cannot say it's a saddle point (we'd need the second order partials to be opposite sign, in this case Z = 0).
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