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Hw15-max-and-min-values: Problem 4 Problem Value: 1 point(s). Problem Score: 61%. Attempts Remaining: 20 attempts. Help Entering Answers (1 point) Part 1 Find and classify all the critical points of $f(x, y) = 5xy(-9 - x - y)$. $f_x = -45y - 10xy - 5y^2$ $f_y = -45x - 5x^2 - 10xy$ $f_{xx} = -10y$ $f_{xy} = -45 - 10x - 10y$ $f_{yy} = -10x$ The critical points of $f$ are: $(0, 0), (-3, -3), (-9, 0), (0, -9)$ (Separate points with commas.) Part 2 $D(0, 0) =$ Therefore $(0, 0)$ is a saddle point $D(-9, 0) =$ Therefore $(-9, 0)$ is a saddle point $D(0, -9) =$ Therefore $(0, -9)$ is a saddle point $D(-3, -3) =$ Therefore $(-3, -3)$ is a local minimum

          Hw15-max-and-min-values: Problem 4
Problem Value: 1 point(s). Problem Score: 61%. Attempts Remaining: 20 attempts.
Help Entering Answers
(1 point)
Part 1
Find and classify all the critical points of $f(x, y) = 5xy(-9 - x - y)$.
$f_x = -45y - 10xy - 5y^2$
$f_y = -45x - 5x^2 - 10xy$
$f_{xx} = -10y$
$f_{xy} = -45 - 10x - 10y$
$f_{yy} = -10x$
The critical points of $f$ are: $(0, 0), (-3, -3), (-9, 0), (0, -9)$
(Separate points with commas.)
Part 2
$D(0, 0) =$
Therefore $(0, 0)$ is a saddle point
$D(-9, 0) =$
Therefore $(-9, 0)$ is a saddle point
$D(0, -9) =$
Therefore $(0, -9)$ is a saddle point
$D(-3, -3) =$
Therefore $(-3, -3)$ is a local minimum

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Hw15-max-and-min-values: Problem 4
Problem Value: 1 point(s). Problem Score: 61%. Attempts Remaining: 20 attempts.
Help Entering Answers
(1 point)
Part 1
Find and classify all the critical points of f(x, y) = 5xy(-9 - x - y).
fx = -45y - 10xy - 5y^2
fy = -45x - 5x^2 - 10xy
fxx = -10y
fxy = -45 - 10x - 10y
fyy = -10x
The critical points of f are: (0, 0), (-3, -3), (-9, 0), (0, -9)
(Separate points with commas.)
Part 2
D(0, 0) =
Therefore (0, 0) is a saddle point
D(-9, 0) =
Therefore (-9, 0) is a saddle point
D(0, -9) =
Therefore (0, -9) is a saddle point
D(-3, -3) =
Therefore (-3, -3) is a local minimum

Added by Sofia H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

07/05/2022

Step 1

Step 1: To find the critical points of the function f(x, y), we need to find the values of x and y where the partial derivatives of f with respect to x and y are equal to zero. Show more…

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Texts: Hw15-max-and-min-values: Problem 4 Problem Value: 1 point(s). Problem Score: 61%. Attempts Remaining: 20 attempts. Help Entering Answers (1 point) Part 1 Find and classify all the critical points of f(x, y) = 5xy(9 - x - y) - 45y - 10xy - 5y^2 - 45x - 5x^2 - 10xy fxx = -10y fyy = -45 - 10x - 10y fy = -10 The critical points of f are: (0,0), (-3,-3), (-9,0), (0,-9) (Separate points with commas.) Part 2 D(0,0) = fxx(0,0)fyy(0,0) - (fxy(0,0))^2 Therefore, D(0,0) = 0 and fxx(0,0) = -10(0) = 0 Therefore, 0,0 is a saddle point of f. D(9, 0) = fxx(9, 0)fyy(9, 0) - (fxy(9, 0))^2 Therefore, D(9, 0) = (-10)(-45 - 10(9) - 10(0)) - (0)^2 = -10(45 - 90) = 450 - 900 = -450 and fxx(-9, 0) = -10(0) = 0 Therefore, 9,0 is a saddle point of f. D(0,9) = fxx(0,9)fyy(0,9) - (fxy(0,9))^2 Therefore, D(0,9) = (-10)(-45 - 10(0) - 10(9)) - (0)^2 = -10(-45 - 90) = 10(135) = 1350 and fxx(0,-9) = -10(-9) = 90 Therefore, 0,-9 is a saddle point of f. D(3, 3) = fxx(3, 3)fyy(3, 3) - (fxy(3, 3))^2 Therefore, D(3, 3) = (-10)(-45 - 10(3) - 10(3)) - (0)^2 = -10(-45 - 60) = 10(105) = 1050 and fx(3, -3) = -10(-3) = 30 Therefore, 3, -3 is a local minimum of f.

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