To determine To find: The dimensions of the rectangular box with largest possible volume. Explanation Given: The rectangular box with an open top and one partition is to be constructed from 288 square inches of cardboard. Result used: Method of Lagrange Multipliers “The below four steps are the procedure for method of Lagrange Multipliers: 1. Rewrite the problem in the form of maximize (or minimize) the function z = f ( x , y ) subject to the constraint g ( x , y ) = 0 . 2. Form the function F : F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) . 3. Compute the critical points of F ; that is, the solution of the system of the equation F x ( x , y , z ) = 0 F y ( x , y , z ) = 0 F z ( x , y , z ) = 0 4. If ( x 0 , y 0 , λ 0 ) is the only critical point of the function F , then the point ( x 0 , y 0 ) is the solution of the given hypothesis. If F has more than one critical point, compute the value of f ( x , y ) at ( x 0 , y 0 ) for each critical point ( x 0 , y 0 , λ 0 ) of F . then assume that the largest value of these values is the maximum value of f ( x , y ) , subject to constraint g ( x , y ) = 0 , and the smallest is the minimum of f ( x , y ) , subject to the constraint g ( x , y ) = 0 .” Calculation: Step 1: Let x × y × z be the dimension of the rectangular box. Then, the volume of the rectangular box is V ( x , y , z ) = x y z . Since the amount of material used is 288 square inches, the x , y and z must satisfy the equation x y + 2 x z + 3 y z = 288 . The constraint of the form g ( x , y , z ) = 0 . x y + 2 x z + 3 y z = 288 x y + 2 x z + 3 y z − 288 = 0 Maximize volume f ( x , y , z ) = x y z Subject to constraint g ( x , y , z ) = x y + 2 x z + 3 y z − 288 . Step 2: Need to form a function F ( x , y , z , λ ) = f ( x , y , z ) + λ g ( x , y , z ) . Substitute f ( x , y , z ) = x y z and g ( x , y , z ) = x y + 2 x z + 3 y z − 288 in F ( x , y , z , λ ) , F ( x , y , z , λ ) = x y z + λ ( x y + 2 x z + 3 y z − 288 ) = x y z + x y λ + 2 x z λ + 3 y z λ − 288 λ Thus, the function F ( x , y , λ ) = x y z + x y λ + 2 x z λ + 3 y z λ − 288 λ . Step 3: In order to find the critical points, first find the partial derivatives. The partial derivative of the function F ( x , y , z , λ ) with respect to x is computed below. F x ( x , y , z , λ ) = ∂ ∂ x ( x y z + x y λ + 2 x z λ + 3 y z λ − 288 λ ) = ∂ ∂ x ( x y z ) + ∂ ∂ x ( x y λ ) + ∂ ∂ x ( 2 x z λ ) + ∂ ∂ x ( 3 y z λ ) − ∂ ∂ x ( 288 λ ) = ( y z ) ( 1 ) + ( y λ ) ( 1 ) + ( 2 z λ ) ( 1 ) + 0 − 0 = y z + y λ + 2 z λ Thus, the partial derivative of F with respect to x is F x ( x , y , z , λ ) = y z + y λ + 2 z λ . The partial derivative of the function F ( x , y , z , λ ) with respect to y is computed below. F y ( x , y , z , λ ) = ∂ ∂ y ( x y z + x y λ + 2 x z λ + 3 y z λ − 288 λ ) = ∂ ∂ y ( x y z ) + ∂ ∂ y ( x y λ ) + ∂ ∂ y ( 2 x z λ ) + ∂ ∂ y ( 3 y z λ ) − ∂ ∂ y ( 288 λ ) = ( x z ) ( 1 ) + ( x λ ) ( 1 ) + 0 + ( 3 z λ ) ( 1 ) − 0 = x z + x λ + 3 z λ Thus, the partial derivative of F with respect to y is F y ( x , y , z , λ ) = x z + x λ + 3 z λ . The partial derivative of the function F ( x , y , z , λ ) with respect to z is computed below. F z ( x , y , z , λ ) = ∂ ∂ z ( x y z + x y λ + 2 x z λ + 3 y z λ − 288 λ ) = ∂ ∂ z ( x y z ) + ∂ ∂ z ( x y λ ) + ∂ ∂ z ( 2 x z λ ) + ∂ ∂ z ( 3 y z λ ) − ∂ ∂ z ( 288 λ ) = ( x y ) ( 1 ) + 0 + ( 2 x λ ) ( 1 ) + ( 3 y λ ) ( 1 ) − 0 = x y + 2 x λ + 3 y λ Thus, the partial derivative of F with respect to z is F z ( x , y , z , λ ) = x y + 2 x λ + 3 y λ

College Mathematics for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134674148

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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