Final Review Guide

# Interpretation when taking the first partial

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* Interpretation: When taking the first partial derivative of f ( x, y ) with respect to x , treat all instances of the variable y (and any true constant) as a constant. Analogous statement with respect to y . See example 1 on p. 542. * Second order partials: see pp. 547-548 for a discussion on second order partials. * Exercises: p.550 3-42 8.3 Maxima and Minima of Functions of Several Variables * Terms: relative maximum, relative maximum value, relative minimum, relative minimum value, absolute maximum, absolute maximum value, absolute minimum, absolute minimum value, saddle point, critical point, second derivative test. * Critical Point of f : A critical point of f is a point ( a, b ) in the domain of f such that both ∂f ∂x ( a, b ) = 0 and ∂f ∂y ( a, b ) = 0 or at least one of the partial derivative does not exist. * Determining Relative Extrema: 1. Find the critical points of f ( x, y ) by solving the system of simultaneous equations f x = 0 f y = 0 2. The second derivative test for the critical point ( a, b ): Let D ( x, y ) = f xx f yy - f 2 xy Then, (a) D ( a, b ) > 0 and f xx ( a, b ) < 0 implies that f ( x, y ) has a relative maximum at the point ( a, b ). (b) D ( a, b ) > 0 and f xx ( a, b ) > 0 implies that f ( x, y ) has a relative minimum at the point ( a, b ). (c) D ( a, b ) < 0 implies that f ( x, y ) has neither a rel. max. nor a rel. min. at the point ( a, b ). The point ( a, b, f ( a, b )) is called a saddle point . (d) D ( a, b ) = 0 implies that the test is inconclusive, so some other technique is needed to solve the problem. * Exercises: pp.561-563 1-30 8.5 Constrained Maxima and Minima and the Method of Lagrange Multipliers * Terms: unconstrained relative extremum, constrained relative extremum, method of Lagrange multipliers, Lagrangian function * Substitution method: Want to find some extremum of a function f ( x, y ) subject to the constraint g ( x, y ) = 0. If g ( x, y ) = 0 is simple enough to solve for x or y , do so. For the sake of argument, say we could solve for y and y = r ( x ) where r ( x ) is some expression in x . Then take the expression for y and plug it into f ( x, y ) giving f ( x, r ( x )). So now you have a function of one variable h ( x ) = f ( x, r ( x )). Now you can use single variable methods to find the desired extremum. This is the same process if you solved for x instead. See example 1 on page 576 for a concrete example of this. 6

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* The Method of Lagrange Multipliers: To find the relative extrema of the function f ( x, y ) subject to the constraint g ( x, y ) = 0 (assuming that these extreme values exist), 1. Form an auxilliary function F ( x, y, λ ) = f ( x, y ) + λg ( x, y ) called the Langrangian function (the variable λ is called the lagrange multiplier). 2. Solve the system that consists of teh equations F x = 0 , F y = 0 , F λ = 0 for all values of x, y, and λ . 3. The solutions found in step 2 are candidates for the extrema of f . * Tip: it is usually a good decision when solving the system in step 2 to first solve F x = 0 and F y = 0 for λ . THen you have two separate expressions for λ . You can then set them equal to one another. This equation along with F λ = 0 will give you two independent equations in x and y . So you have two equations and two unknowns. You can then finish the system. In the grand scheme of the problem, it is never important what the value(s) of λ
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