(1) Lagrange multiplier is a very useful technique to determine the extremum of a function under constraint(s). For a function f (x, y), the extremum (minimum or maximum) is determined by evaluating af /dx = af /@y = 0. But determining the extremum under a constraint g(x, y) = 0 can be tricky. Even if y can be expressed explicitly in terms of x from g(x, y) = 0, the calculations may become tedious. Here's Lagrange multiplier method: For a function f (x, y), the extremum under a constraint g(x, y) = 0 is determined by defining L(x,y,2) = f(x,y) – Ag(x, y) and by evaluating ôL/dx = aL/ðy = aL/ð1 = 0. Using this method, determine the maximum value of f(x,y) = 2x + 3y under the constraint x² + y?. = 1 (hint: you need to check which extremum gives the maximum value).
(1) Lagrange multiplier is a very useful technique to determine the extremum of a function under constraint(s). For a function f (x, y), the extremum (minimum or maximum) is determined by evaluating af /dx = af /@y = 0. But determining the extremum under a constraint g(x, y) = 0 can be tricky. Even if y can be expressed explicitly in terms of x from g(x, y) = 0, the calculations may become tedious. Here's Lagrange multiplier method: For a function f (x, y), the extremum under a constraint g(x, y) = 0 is determined by defining L(x,y,2) = f(x,y) – Ag(x, y) and by evaluating ôL/dx = aL/ðy = aL/ð1 = 0. Using this method, determine the maximum value of f(x,y) = 2x + 3y under the constraint x² + y?. = 1 (hint: you need to check which extremum gives the maximum value).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:(1) Lagrange multiplier is a very useful technique to determine the extremum of a function under
constraint(s). For a function f (x, y), the extremum (minimum or maximum) is determined by
evaluating ôf /dx
can be tricky. Even if y can be expressed explicitly in terms of x from g(x, y) = 0, the calculations
may become tedious.
af/ðy = 0. But determining the extremum under a constraint g(x, y) = 0
Here's Lagrange multiplier method:
For a function f (x,y), the extremum under a constraint g(x, y) = 0 is determined by defining
L(x, y, 1) = f(x, y) – Ag(x, y) and by evaluating ôL/əx = aL/@y = aL/a = 0.
%3D
Using this method, determine the maximum value of f(x, y) = 2x + 3y under the constraint
x2 + y?
= 1 (hint: you need to check which extremum gives the maximum value).
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