Explanation Given: The sum of the length and girth of the mailing carton rectangular package cannot exceed 120 inches. Calculation: Let x , y and z be the width, height and length of the rectangular package, respectively. Then, the sum of the girth and length of the rectangular package is 2 x + 2 y + z . It is known that, the sum of the girth and length of the rectangular package is 120. That is, 2 x + 2 y + z = 120 The volume of the rectangular package is equal to product of length and breadth and height. That is, the volume V ( x , y , z ) = x y z . The constrain of the form is g ( x , y ) = 0 . 2 x + 2 y + z = 120 2 x + 2 y + z − 120 = 0 Maximize f ( x , y ) = x y z subject to the constraint g ( x , y ) = 2 x + 2 y + z − 120 . Need to form a function F ( x , y , λ ) = f ( x , y ) + λ g ( x , y ) . Substitute f ( x , y , z ) = x y z and g ( x , y , z ) = 2 x + 2 y + z − 120 in F ( x , y , z , λ ) , F ( x , y , z , λ ) = x y z + λ ( 2 x + 2 y + z − 120 ) = x y z + 2 x λ + 2 y λ + z λ − 120 λ Thus, the function F ( x , y , z , λ ) = x y z + 2 x λ + 2 y λ + z λ − 120 λ . 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 , λ ) = ∂ ∂ x ( x y z + 2 x λ + 2 y λ + z λ − 120 λ ) = ∂ ∂ x ( x y z ) + ∂ ∂ x ( 2 x λ ) + ∂ ∂ x ( 2 y λ ) + ∂ ∂ x ( z λ ) − ∂ ∂ x ( 120 λ ) = ( y z ) ∂ ∂ x ( x ) + ( 2 λ ) ∂ ∂ x ( x ) + 0 + 0 − 0 = y z + 2 λ Therefore, the partial derivative of F with respect to x is F x ( x , y , λ ) = y z + 2 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 + 2 x λ + 2 y λ + z λ − 120 λ ) = ∂ ∂ y ( x y z ) + ∂ ∂ y ( 2 x λ ) + ∂ ∂ y ( 2 y λ ) + ∂ ∂ y ( z λ ) − ∂ ∂ y ( 120 λ ) = ( x z ) ∂ ∂ y ( y ) + 0 + ( 2 λ ) ∂ ∂ y ( y ) + 0 − 0 = x z + 2 λ Thus, the partial derivative of the function F with respect to y F y ( x , y , z , λ ) = x z + 2 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 + 2 x λ + 2 y λ + z λ − 120 λ ) = ∂ ∂ z ( x y z ) + ∂ ∂ z ( 2 x λ ) + ∂ ∂ z ( 2 y λ ) + ∂ ∂ z ( z λ ) − ∂ ∂ z ( 120 λ ) = ( x y ) ∂ ∂ z ( z ) + 0 + 0 + ( λ ) ∂ ∂ z ( z ) − 0 = x y + λ Thus, the partial derivative of the function F with respect to z is F z ( x , y , z , λ ) = x y + λ

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Chapter 7.4, Problem 36E

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Chapter 7.4, Problem 35E

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