Explanation Given Information: The provided function is, z = 6 x 2 + x 2 y + y 2 − 4 y + 9 . Formula used: To calculate relative maxima and minima of the z = f ( x , y ) , 1. Find the partial derivatives ∂ z ∂ x and ∂ z ∂ y . 2. Find the critical points, that is, the point(s) that satisfy ∂ z ∂ x = 0 and ∂ z ∂ y = 0 . 3. Then find all the second partial derivatives and evaluate the value of D at each critical point, where D = ( z x x ) ( z y y ) − ( z x y ) 2 = ∂ 2 z ∂ x 2 ⋅ ∂ 2 z ∂ y 2 − ( ∂ 2 z ∂ x ∂ y ) 2 . (a) If D > 0 , then relative minimum occurs if z x x > 0 and relative maximum occurs if z x x < 0 . (b) If D < 0 , then neither a relative maximum nor a relative minimum occurs. For a function f ( x , y ) , the partial derivative of f with respect to x is calculated by taking the derivative of f ( x , y ) with respect to x and keeping the other variable y constant and the partial derivative of f with respect to y is calculated by taking the derivative of f ( x , y ) with respect to y and keeping the other variable x constant. The partial derivative of f with respect to x is denoted by f x and with respect to y is denoted by f y . For a function z ( x , y ) , the second partial derivative, 1. When both derivatives are taken with respect to x is z x x = ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) . 2. When both derivatives are taken with respect to y is z y y = ∂ 2 z ∂ y 2 = ∂ ∂ y ( ∂ z ∂ y ) . 3. When first derivative is taken with respect to x and second derivative is taken with respect to y is z x y = ∂ 2 z ∂ y ∂ x = ∂ ∂ y ( ∂ z ∂ x ) . 4. When first derivative is taken with respect to y and second derivative is taken with respect to x is z y x = ∂ 2 z ∂ x ∂ y = ∂ ∂ x ( ∂ z ∂ y ) . Power of x rule for a real number n is such that, if f ( x ) = x n then f ′ ( x ) = n x n − 1 . Chain rule for function f ( x ) = u ( v ( x ) ) is f ′ ( x ) = u ′ ( v ( x ) ) ⋅ v ′ ( x ) . Constant function rule for a constant c is such that, if f ( x ) = c then f ′ ( x ) = 0 . Coefficient rule for a constant c is such that, if f ( x ) = c ⋅ u ( x ) , where u ( x ) is a differentiable function of x , then f ′ ( x ) = c ⋅ u ′ ( x ) . Calculation: Consider the function, z = 6 x 2 + x 2 y + y 2 − 4 y + 9 . Recall that, for a function f ( x , y ) , the partial derivative of f with respect to x is calculated by taking the derivative of f ( x , y ) with respect to x and keeping the other variable y constant and the partial derivative of f with respect to y is calculated by taking the derivative of f ( x , y ) with respect to y and keeping the other variable x constant. Use the power of x rule for derivatives, the constant function rule, the chain rule, and the coefficient rule, Thus, ∂ z ∂ x = 0 12 x + 2 x y = 0 2 x ( 6 + y ) = 0 Thus, x = 0 or y = − 6 . And, ∂ z ∂ y = 0 x 2 + 2 y − 4 = 0 Consider the equation, x = 0 Substitute 0 for x in x 2 + 2 y − 4 = 0 . 0 + 2 y − 4 = 0 y = 2 Consider the equation, y = − 6 Substitute − 6 for y in x 2 + 2 y − 4 = 0 . x 2 + 2 ( − 6 ) − 4 = 0 x 2 − 12 − 4 = 0 x 2 = 16 x = ± 4 Thus, the critical points are ( 0 , 2 ) , ( 4 , − 6 ) and ( − 4 , − 6 ) . Recall that, for a function z ( x , y ) , the second partial derivative, when both derivatives are taken with respect to x is z x x = ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) , when both derivatives are taken with respect to y is z y y = ∂ 2 z ∂ y 2 = ∂ ∂ y ( ∂ z ∂ y ) , when first derivative is taken with respect to x and second derivative is taken with respect to y is z x y = ∂ 2 z ∂ y ∂ x = ∂ ∂ y ( ∂ z ∂ x )

MATHMATICAL APPLICATIONS FOR THE MANAGEM

11th Edition

ISBN: 9781305758063

Author: HARSHBARGER

Publisher: CENGAGE C

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Question

Chapter 14, Problem 3T

To determine

To calculate: The pairs (x,y) that are critical points for z and then classify each as a relative maxima, relative minima, and saddle points of z=6x2+x2y+y2−4y+9.

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Chapter 14, Problem 2T

Chapter 14, Problem 4T

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