Explanation Given: The function is f ( x , y ) = x y 2 + x y (1) Result used: Second derivative test “Let D ( x , y ) = f x x ( x , y ) f y y ( x , y ) − f 2 x y ( x , y ) . Then, check the values of D ( a , b ) as below. (a) If D ( a , b ) > 0 and f x x < 0 then the function f has a relative maximum at the critical point ( a , b ) . (b) If D ( a , b ) > 0 and f x x > 0 then the function f has a relative minimum at the critical point ( a , b ) . (c) If D ( a , b ) < 0 then the function f has a no relative extrema at the critical point ( a , b ) . The point ( a , b , f ( a , b ) ) is said to be a saddle point. (d) If D ( a , b ) = 0 then there will be no result.” Calculation: Obtain the critical point of the function . The partial derivatives of the given function is computed as follows, Partially differentiate the equation (1) with respect to x , ∂ ∂ x f ( x , y ) = ∂ ∂ x ( x y 2 + x y ) ∂ ∂ x f ( x , y ) = 1 y 2 + y (2) Partially differentiate the equation (1) with respect to y , ∂ ∂ y f ( x , y ) = ∂ ∂ y ( x y 2 + x y ) ∂ ∂ y f ( x , y ) = − 2 x y 3 + x (3) Thus, the partial derivatives are f x = 1 y 2 + y and f y = − 2 x y 3 + x . Find the critical points by setting f x ( x , y ) = 0 and f y ( x , y ) = 0 . The equation (2) can be written as follows, 1 y 2 + y = 0 1 y 2 + y = 0 1 + y 3 y 2 = 0 1 + y 3 = 0 (4) The equation (3) can be written as follows, − 2 x y 3 + x = 0 − 2 x + x y 3 y 3 = 0 − 2 x + x y 3 = 0 (5) Solve the above equations (4) and (5). Obtain the value of y from equation (4). 1 + y 3 = 0 ( y + 1 ) ( y 2 − y + 1 ) = 0 Consider ( y + 1 ) = 0 , then y = − 1 Use the quadratic formula to solve the equation y 2 − y + 1 = 0 , x = − b ± b 2 − 4 a c 2 a (6) Substitute a = 1 , b = − 1 and c = 1 in the equation (6), y = 1 ± ( − 1 ) 2 − 4 ( 1 ) ( 1 ) 2 ( 1 ) = 1 ± 1 − 4 2 y = 1 ± i 3 2 Thus, the values of y are y = − 1 and y = 1 ± i 3 2 Multiply the equation (4) by x , x + x y 3 = 0 (7) Subtract the equation (7) from the equation (5)

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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ISBN: 9780357465400

Author: Tan

Publisher: CENGAGE L

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Chapter 8.3, Problem 14E

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The relative extrema of the function by finding the critical points and classify the nature of each point by using second derivative test.

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According to Newton's law of universal gravitation, the force F between two bodies of constant mass GmM m and M is given by the formula F = , where G is the gravitational constant and d is the d² distance between the bodies. a. Suppose that G, m, and M are constants. Find the rate of change of force F with respect to distance d. F' (d) 2GmM b. Find the rate of change of force F with gravitational constant G = 6.67 × 10-¹¹ Nm²/kg², on two bodies 5 meters apart, each with a mass of 250 kilograms. Answer in scientific notation, rounding to 2 decimal places. -6.67x10 N/m syntax incomplete.

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APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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Ch. 8.1 - Prob. 8E Ch. 8.1 - Let g(r, s, t) = res/t. 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The relative extrema of the function by finding the critical points and classify the nature of each point by using second derivative test.

Solution Summary: The author explains the relative extrema of the function by finding the critical points and classifying the nature of each point by using second derivative test.

The relative extrema of the function by finding the critical points and classify the nature of each point by using second derivative test.

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Chapter 8 Solutions