Explanation Formula used: Lagrange Multipliers: To obtain the relative extremum of the function f ( x , y , z ) subject to the constraint g ( x , y , z ) = 0 (assuming that these extreme values exist), (1) Form an auxiliary function F ( x , y , z , λ ) = f ( x , y , z ) + λ g ( x , y , z ) called the Lagrangian function where the variable λ is called the Lagrange multiplier. (2) Solve the system that consists of the equations F x = 0 , F y = 0 , F z = 0 and F λ = 0 for all values of x , y , z and λ . (3) The solutions found in step (2) are candidates for the extremum of f . Calculation: Consider the function g ( x , y , z ) = 4 x 2 + y 2 − z . Given that 4 x 2 + y 2 − z . hence, obtain the value of y explicitly in terms of x from the function is g ( x , y , z ) = 0 . The auxiliary function is, F ( x , y , z , λ ) = f ( x , y , z ) + λ g ( x , y , z ) = x + 2 y − 3 z + λ ( 4 x 2 + y 2 − z ) The above expression is called the Lagrange function while λ be the Lagrange multipliers. Compute the term F x from the Lagrangian function as given below. Differentiate F ( x , y , z , λ ) = x + 2 y − 3 z + λ ( 4 x 2 + y 2 − z ) with respect to x , F x = ∂ ∂ x ( x + 2 y − 3 z + λ ( 4 x 2 + y 2 − z ) ) = ∂ ∂ x ( x + 2 y − 3 z ) + ∂ ∂ x ( λ ( 4 x 2 + y 2 − z ) ) = 1 + 8 x λ By the step (2) of the above formula it is observed that 1 + 8 x λ = 0 . Compute the term F y from the Lagrangian function. Differentiate F ( x , y , z , λ ) = x + 2 y − 3 z + λ ( 4 x 2 + y 2 − z ) with respect to y , F y = ∂ ∂ y ( x + 2 y − 3 z + λ ( 4 x 2 + y 2 − z ) ) = ∂ ∂ y ( x + 2 y − 3 z ) + ∂ ∂ y ( λ ( 4 x 2 + y 2 − z ) ) = 2 + λ ( 2 y ) = 2 + 2 y λ By the step (2) of the above formula it is observed that 2 + 2 y λ = 0 . Compute the term F z from the Lagrangian function. Differentiate F ( x , y , z , λ ) = x + 2 y − 3 z + λ ( 4 x 2 + y 2 − z ) with respect to z , F z = ∂ ∂ z ( x + 2 y − 3 z + λ ( 4 x 2 + y 2 − z ) ) = ∂ ∂ z ( x + 2 y − 3 z ) + ∂ ∂ z ( λ ( 4 x 2 + y 2 − z ) ) = − 3 + λ ( − 1 ) = − 3 − λ By the step (2) of the above formula it is observed that − 3 − λ = 0

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Chapter 8.5, Problem 16E

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To maximize: The function f(x,y,z)=x+2y−3z subjects to the constraint of function z=4x2+y2 .

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Chapter 8.5, Problem 15E

Chapter 8.5, Problem 17E

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To maximize: The function f ( x , y , z ) = x + 2 y − 3 z subjects to the constraint of function z = 4 x 2 + y 2 .

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To maximize: The function f(x,y,z)=x+2y−3z subjects to the constraint of function z=4x2+y2 .

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