Explanation Given Information: The provided function is S = x y + x 2 z 2 + 4 y z subject to x y z = 1 . Formula used: Steps to solve a maxima - minima problem using substitution, Step 1: Solve the constraint equation for one of the variables, and then substitute in the objective function. Step 2: Minimize/Maximize the resulting function of two variables. The derivative formula, d d x ( x n ) = n x n − 1 Where, n is any real number. Second derivative test for two variable function, If ( a , b ) is a critical point of any function f of two variables, then H = f x x ( a , b ) f y y ( a , b ) − [ f x y ( a , b ) ] 2 Where, H be a mathematical quantity. If, H is positive, then f has relative minimum at ( a , b ) if f x x ( a , b ) > 0 and f has relative maximum at ( a , b ) if f x x ( a , b ) < 0 . And, if H is negative, then f has a saddle point at ( a , b ) . Calculation: Consider the provided function, S = x y + x 2 z 2 + 4 y z Now, solve the constraint equation for one of the variables, x y z = 1 z = 1 x y …… (1) Thus, substitute 1 x y for z in the objective function as, S = x y + x 2 ( 1 x y ) 2 + 4 y ( 1 x y ) = x y + 1 y 2 + 4 x Now, minimize the resulting function of two variables. First obtain the partial derivative with respect to x as, S x = ∂ ∂ x ( x y + 1 y 2 + 4 x ) = ∂ ∂ x ( x y ) + ∂ ∂ x ( 1 y 2 ) + ∂ ∂ x ( 4 x ) = y − 4 x 2 Obtain critical point by setting S x = 0 as, y − 4 x 2 = 0 y = 4 x 2 Now, obtain the derivative with respect to y as, S y = ∂ ∂ y ( x y + 1 y 2 + 4 x ) = ∂ ∂ y ( x y ) + ∂ ∂ y ( 1 y 2 ) + ∂ ∂ y ( 4 x ) = x − 2 y 3 Set S y = 0 , x − 2 y 3 = 0 Substitute 4 x 2 for y as, x − 2 ( 4 x 2 ) 3 = 0 x − x 6 32 = 0 x ( 1 − x 5 32 ) = 0 Thus, x = 0 And, 1 − x 5 32 = 0 x 5 32 = 1 x 5 = 32 x 5 = 2 5 Thus, x = 2

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Author: Waner, Stefan.

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Chapter 8, Problem 30RE

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To calculate: The minimum value of S=xy+x2z2+4yz subject to xyz=1 where x>0, y>0 and z>0.

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Chapter 8, Problem 29RE

Chapter 8, Problem 31RE

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The minimum value of S = x y + x 2 z 2 + 4 y z subject to x y z = 1 where x > 0 , y > 0 and z > 0 .

Solution Summary: The author explains the formula used to solve a maxima-minima problem using substitution.

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To calculate: The minimum value of S=xy+x2z2+4yz subject to xyz=1 where x>0, y>0 and z>0.

Want to see the full answer?

Chapter 8 Solutions

The minimum value of S = x y + x 2 z 2 + 4 y z subject to x y z = 1 where x &gt; 0 , y &gt; 0 and z &gt; 0 .

Solution Summary: The author explains the formula used to solve a maxima-minima problem using substitution.

Videos

To calculate: The minimum value of S=xy+x2z2+4yz subject to xyz=1 where x>0, y>0 and z>0.

Want to see the full answer?

Chapter 8 Solutions

The minimum value of S = x y + x 2 z 2 + 4 y z subject to x y z = 1 where x > 0 , y > 0 and z > 0 .