Difference between revisions of "Lagrange multiplier"

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(better, with formula)
(example worked out)
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and ''g(x,y) = 0''.
 
and ''g(x,y) = 0''.
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== Example ==
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''Problem'': What is the shape and dimension of the largest rectangle that fits within a circle having a radius equal to the square root of 2?
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''Solution'':  ''g(x,y) = x<sup>2</sup> + y<sup>2</sup> - 2 = 0''.  We want to maximize the area of a rectangle.  Assuming symmetry about the origin (we could always map a solution to a symmetric one), the area ''f(x,y)'' that we want to maximize equals 2x times 2y:  ''f(x,y) = 4xy''.  From the equation for the Lagrangian multiplier above, and setting the ''x'' and ''y'' coefficients equal to each other, we can generate these two equations:
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:4y = \lambda 2x
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:4x = \lambda 2y
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Solving each for lambda (λ) demonstrates that x must be equal to y, and plugging that into the equation for ''g'' yields:
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:2x<sup>2</sup> = 2
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:x = 1 = y
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Then do not forget to include the maximum area in your answer, which is ''f(1,1) = 4(1)(1) = 4.
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[[Category:calculus]]
 
[[Category:calculus]]
 
[[Category:mathematics]]
 
[[Category:mathematics]]

Revision as of 23:52, December 25, 2009

A Lagrangian multiplier is a constant, usually represented by the Greek symbol lambda (λ), which is part of the formula for finding an extreme value of a function f(x,y) subject to a constraint g(x,y). For example, the method of Lagrange multipliers can help determine the maximum area of a rectangle that can be fit inside a given circle represented by g(x,y) = x2 + y2.

The points at which extrema occur for a function f(x,y) subject to the constraint g(x,y) are the common solutions to:

and g(x,y) = 0.

Example

Problem: What is the shape and dimension of the largest rectangle that fits within a circle having a radius equal to the square root of 2?

Solution: g(x,y) = x2 + y2 - 2 = 0. We want to maximize the area of a rectangle. Assuming symmetry about the origin (we could always map a solution to a symmetric one), the area f(x,y) that we want to maximize equals 2x times 2y: f(x,y) = 4xy. From the equation for the Lagrangian multiplier above, and setting the x and y coefficients equal to each other, we can generate these two equations:

4y = \lambda 2x
4x = \lambda 2y

Solving each for lambda (λ) demonstrates that x must be equal to y, and plugging that into the equation for g yields:

2x2 = 2
x = 1 = y

Then do not forget to include the maximum area in your answer, which is f(1,1) = 4(1)(1) = 4.