Explanation Given information: The function is f ( x , y ) = − 5 x 2 + 4 x y − 17 y 2 − 6 x + 6 y + 2 . Explanation: If f ( x , y ) has either a relative maximum or minimum at point ( x , y ) = ( a , b ) , then ∂ f ∂ x ( a , b ) = 0 and ∂ f ∂ y ( a , b ) = 0 . Differentiating f ( x , y ) = − 5 x 2 + 4 x y − 17 y 2 − 6 x + 6 y + 2 partially with respect to x and y , ∂ f ∂ x = ∂ ∂ x ( − 5 x 2 + 4 x y − 17 y 2 − 6 x + 6 y + 2 ) ⇒ ∂ f ∂ x = − 10 x + 4 y − 6 ∂ f ∂ y = ∂ ∂ y ( − 5 x 2 + 4 x y − 17 y 2 − 6 x + 6 y + 2 ) ⇒ ∂ f ∂ y = 4 x − 34 y + 6 . Now, set ∂ f ∂ x = 0 and ∂ f ∂ y = 0 , and solve for x and y . Therefore, − 10 x + 4 y − 6 = 0 and 4 x − 34 y + 6 = 0 . By solving the equation − 10 x + 4 y − 6 = 0 for y . ⇒ 4 y = 10 x + 6 ⇒ y = 5 2 x + 3 2 By substituting y = 5 2 x + 3 2 in 4 x − 34 y + 6 = 0 , ⇒ 4 x − 34 ( 5 2 x + 3 2 ) + 6 = 0 By distributing − 34 into parenthesis, ⇒ 4 x − 85 x − 51 + 6 = 0 By combining the like terms, ⇒ − 81 x − 45 = 0 ⇒ 81 x = − 45 By dividing both the sides by 81 , ⇒ x = − 45 81 ⇒ x = − 5 9 Substitute x = − 5 9 in y = 5 2 x + 3 2 and simplify. ⇒ y = 5 2 ( − 5 9 ) + 3 2 ⇒ y = − 25 18 + 3 2 = − 25 + 27 18 ⇒ y = 2 18 = 1 9 Therefore, the function f ( x , y ) = − 5 x 2 + 4 x y − 17 y 2 − 6 x + 6 y + 2 has a possible relative maximum or minimum at ( − 5 9 , 1 9 ) . Now, by the second derivative test there are four conditions to determine a relative maximum and minimum. f ( x , y ) has a relative minimum at ( a , b ) , if D ( a , b ) > 0 and ∂ 2 f ∂ x 2 ( a , b ) > 0 . f ( x , y ) has a relative maximum at ( a , b ) , if D ( a , b ) > 0 and ∂ 2 f ∂ x 2 ( a , b ) < 0 . f ( x , y ) has neither a relative maximum nor a relative minimum at ( a , b ) , if D ( a , b ) < 0

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ISBN: 9780137590810

Author: Larry J. Goldstein, David C. Lay, David I. Schneider, Nakhle H. Asmar, William Edward Tavernetti

Publisher: PEARSON

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Chapter 7.3, Problem 25E

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All the points (x, y) where the function f(x, y)=−5x2+4xy−17y2−6x+6y+2 has a possible relative maximum or minimum, and the nature of f(x, y) by using second-derivative test. If the second-derivative test is inconclusive, then state it.

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Chapter 7.3, Problem 24E

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Question 1: If a barometer were built using oil (p = 0.92 g/cm³) instead of mercury (p = 13.6 g/cm³), would the column of oil be higher than, lower than, or the same as the column of mercury at 1.00 atm? If the level is different, by what factor? Explain. (5 pts) Solution: A barometer works based on the principle that the pressure exerted by the liquid column balances atmospheric pressure. The pressure is given by: P = pgh Since the atmospheric pressure remains constant (P = 1.00 atm), the height of the liquid column is inversely proportional to its density: Step 1: Given Data PHg hol=hgx Poil • Density of mercury: PHg = 13.6 g/cm³ Density of oil: Poil = 0.92 g/cm³ • Standard height of mercury at 1.00 atm: hμg Step 2: Compute Height of Oil = 760 mm = 0.760 m 13.6 hoil = 0.760 x 0.92 hoil = 0.760 × 14.78 hoil = 11.23 m Step 3: Compare Heights Since oil is less dense than mercury, the column of oil must be much taller than that of mercury. The factor by which it is taller is: Final…

Chapter 7 Solutions

Calculus & Its Applications

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All the points ( x , y ) where the function f ( x , y ) = − 5 x 2 + 4 x y − 17 y 2 − 6 x + 6 y + 2 has a possible relative maximum or minimum, and the nature of f ( x , y ) by using second-derivative test. If the second-derivative test is inconclusive, then state it.

Solution Summary: The author explains that the function f(x,y) has a possible relative maximum or minimum at the point . If the second-derivative test is inconclusive, then state it

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All the points (x, y) where the function f(x, y)=−5x2+4xy−17y2−6x+6y+2 has a possible relative maximum or minimum, and the nature of f(x, y) by using second-derivative test. If the second-derivative test is inconclusive, then state it.

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