Explanation Given information: The function is f ( x , y ) = − 2 x 2 + 2 x y − y 2 + 4 x − 6 y + 5 . 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 . Therefore, find the partial derivatives ∂ f ∂ x and ∂ f ∂ y of the given function. By differentiating the given function with respect to x , s ∂ f ∂ x = ∂ ∂ x ( − 2 x 2 + 2 x y − y 2 + 4 x − 6 y + 5 ) ⇒ ∂ f ∂ x = − 4 x + 2 y + 4 . Now, differentiate the given function with respect to y . ∂ f ∂ y = ∂ ∂ y ( − 2 x 2 + 2 x y − y 2 + 4 x − 6 y + 5 ) ⇒ ∂ f ∂ y = 2 x − 2 y − 6 . Now, set ∂ f ∂ x = 0 and ∂ f ∂ y = 0 and solve for x and y . Therefore, − 4 x + 2 y + 4 = 0 and 2 x − 2 y − 6 = 0 . By solving the equation − 4 x + 2 y + 4 = 0 for y . ⇒ 2 y = 4 x − 4 ⇒ y = 2 x − 2 By substituting y = 2 x − 2 in 2 x − 2 y − 6 = 0 , ⇒ 2 x − 2 ( 2 x − 2 ) − 6 = 0 By distributing − 2 in to parenthesis, ⇒ 2 x − 4 x + 4 − 6 = 0 By combining the like terms, ⇒ − 2 x − 2 = 0 ⇒ 2 x = − 2 ⇒ x = − 1 Substitute x = − 1 in y = 2 x − 2 and simplify. ⇒ y = 2 ( − 1 ) − 2 ⇒ y = − 4 . Therefore, the function f ( x , y ) = − 2 x 2 + 2 x y − y 2 + 4 x − 6 y + 5 has a possible relative maximum or minimum at ( − 1 , − 4 ) . Now, by the second derivative test there are four conditions to determine a relative maximum and minimum. A point ( a , b ) is relative minimum of f ( x , y ) , if D ( a , b ) > 0 and ∂ 2 f ∂ x 2 ( a , b ) > 0 . A point ( a , b ) is relative maximum of f ( x , y ) , if D ( a , b ) > 0 and ∂ 2 f ∂ x 2 ( a , b ) < 0 . There is no relative maximum or minimum of f ( x , y ) at point ( a , b ) , if D ( a , b ) < 0 . If D ( a , b ) = 0 , then no conclusion can be drawn from this test, where D ( x , y ) = ∂ 2 f ∂ x 2 × ∂ 2 f ∂ y 2 − ( ∂ 2 f ∂ x ∂ y ) 2

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Chapter 7.3, Problem 39E

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All the points (x, y) where the function f(x, y)=−2x2+2xy−y2+4x−6y+5 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 38E

Chapter 7.3, Problem 40E

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Ch. 7.1 - Let f(x,y,z)=x2+y(xz)4. Compute f(3,5,2).Ch. 7.1 - Prob. 2CYU Ch. 7.1 - Let f(x,y)=x23xyy2. Compute f(5,0),f(5,2) and...Ch. 7.1 - Prob. 2E Ch. 7.1 - Let g(x,y,z)=x/(yz). Compute g(1,1),g(0,1) and...Ch. 7.1 - Let f(x,y,z)=x2ey2+z2. Compute f(1,1,1) and...Ch. 7.1 - Let f(x,y)=xy. Show that f(2+h,3)f(2,3)=3h.Ch. 7.1 - Let f(x,y)=xy. Show that f(2,3+k)f(2,3)=2k.Ch. 7.1 - Cost Find a formula C(x,y,z) that gives the cost...Ch. 7.1 - Cost Find a formula C(x,y,z) that gives the cost...

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All the points ( x , y ) where the function f ( x , y ) = − 2 x 2 + 2 x y − y 2 + 4 x − 6 y + 5 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

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All the points (x, y) where the function f(x, y)=−2x2+2xy−y2+4x−6y+5 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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