Try the fastest way to create flashcards
Question

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema:

Evaluate f at each of the solution points and select the extreme value subject to the constraints asked for in the exercise.

Minimize f(x,y,z)=x2+y2+z2f ( x , y , z ) = x ^ { 2 } + y ^ { 2 } + z ^ { 2 } subject to the constraints x2xy+y2z21=0 and x2+y21=0.x ^ { 2 } - x y + y ^ { 2 } - z ^ { 2 } - 1 = 0 \text { and } x ^ { 2 } + y ^ { 2 } - 1 = 0.

Solution

Verified
Answered 9 months ago
Answered 9 months ago
Step 1
1 of 3

All that remains to be done is to evaluate f(x,y,z)f(x,y,z) at the points we obtained from the previous part of the exercise and select the minimal value. For the evaluations of ff we have:

f(1,0,0)=1+0+0=1,f(0,1,0)=0+1+0=1f(0,1,0)=0+1+0=1f(1,0,0)=1+0+0=1f(12,12,12)=12+12+12=32f(12,12,12)=12+12+12=32f(12,12,12)=12+12+12=32f(12,12,12)=12+12+12=32\begin{aligned} & f(-1,0,0) = 1 +0+0=1, \\ & f(0,-1,0) =0+1 +0=1\\ & f(0,1,0) = 0+1+0=1 \\ & f(1,0,0) = 1 +0+0=1 \\ & f\left( - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \\ & f\left( - \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \\ & f\left( \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \\ & f\left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \end{aligned}

From these values we see that the minimal value of ff under the given constraints is 1 and is reached at the points (1,0,0),(0,1,0),(0,1,0),(1,0,0)(-1,0,0),(0,-1,0),(0,1,0),(1,0,0).

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Thomas' Calculus 14th Edition by Christopher E Heil, Joel R. Hass, Maurice D. Weir

Thomas' Calculus

14th EditionISBN: 9780134438986 (11 more)Christopher E Heil, Joel R. Hass, Maurice D. Weir
10,142 solutions
Calculus: Early Transcendentals 8th Edition by James Stewart

Calculus: Early Transcendentals

8th EditionISBN: 9781285741550 (4 more)James Stewart
11,085 solutions
Calculus: Early Transcendentals 9th Edition by Daniel K. Clegg, James Stewart, Saleem Watson

Calculus: Early Transcendentals

9th EditionISBN: 9781337613927 (1 more)Daniel K. Clegg, James Stewart, Saleem Watson
11,049 solutions
Calculus 9th Edition by Daniel K. Clegg, James Stewart, Saleem Watson

Calculus

9th EditionISBN: 9781337624183 (1 more)Daniel K. Clegg, James Stewart, Saleem Watson
10,873 solutions

More related questions

1/4

1/7