Question 1

Use the given tabular representation of the function  fff  to compute  f(2,4) −f(4,2) f ( 2,4 )  - f ( 4,2 ) f(2,4) −f(4,2)   .    x2468y2−5197185−54−9191100−46−20201225−198−19216235−18\begin{array} { | l | l | l | l | l | l | } \hline & & \boldsymbol { x } & & & \\hline & & \mathbf { 2 } & \mathbf { 4 } & \mathbf { 6 } & \mathbf { 8 } \\hline \boldsymbol { y } & \mathbf { 2 } & - 5 & 197 & 185 & - 5 \\hline & \mathbf { 4 } & - 9 & 191 & 100 & - 4 \\hline & \mathbf { 6 } & - 20 & 201 & 225 & - 19 \\hline & \mathbf { 8 } & - 19 & 216 & 235 & - 18 \\hline\end{array}y​2468​x2−5−9−20−19​4197191201216​6185100225235​8−5−4−19−18​​

Accepted Answer

A)  -206
B)   188
C)   197
D)   206
E)   -9

Question 2

Use Lagrange Multipliers to solve the given problem.  
Find the maximum value of  f(x,y) =xyf ( x , y )  = x yf(x,y) =xy  subject to  2x+y=602 x + y = 602x+y=60  .

Accepted Answer

A)    fmax⁡=10f _ { \max } = 10fmax​=10 
B)    fmax⁡=450f _ { \max } = 450fmax​=450 
C)    fmax⁡=30f _ { \max } = 30fmax​=30 
D)    fmax⁡=2f _ { \max } = 2fmax​=2 
E)    fmax⁡=225f _ { \max } = 225fmax​=225

Question 3

If positive electric charges of Q and q coulombs are situated at positions $( a , b , c )$ and $( x , y , z )$ , respectively, then the force of repulsion they experience is given by 
  $F = K \frac { Q q } { ( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } + ( z - c ) ^ { 2 } }$ 
Where $K \approx 9 \times 10 ^ { 9 }$ , F is given in newtons, and all positions are measured in meters. Assume that a charge of 4 coulombs is situated at the origin and that a second charge of 14 coulombs is situated $( 1,4,3 )$ and moving in the y direction at 1.4 meters per second. How fast is the electrostatic force it experiences decreasing 
 
NOTE: Please enter your answer without the units. Round the answer to the nearest trillion.

Accepted Answer

The answer of If positive electric charges of Q and...

Question 4

Evaluate  f(a,2) f ( a , 2 ) f(a,2)   for    f(x,y) =x2−y−xy+5f ( x , y )  = x ^ { 2 } - y - x y + 5f(x,y) =x2−y−xy+5

Accepted Answer

A)    a2−2aa ^ { 2 } - 2 aa2−2a 
B)    a2−2a+3a ^ { 2 } - 2 a + 3a2−2a+3 
C)    a2+2a−2a ^ { 2 } + 2 a - 2a2+2a−2 
D)    a2−2a−2a ^ { 2 } - 2 a - 2a2−2a−2 
E)    a2−2a+5a ^ { 2 } - 2 a + 5a2−2a+5

Question 5

Locate the local maximum of the function.    f(x,y) =x2y−8x2−4y2f ( x , y )  = x ^ { 2 } y - 8 x ^ { 2 } - 4 y ^ { 2 }f(x,y) =x2y−8x2−4y2

Accepted Answer

A)    (8,4,−256) ( 8,4 , - 256 ) (8,4,−256)  
B)    (0,0,0) ( 0,0,0 ) (0,0,0)  
C)    (−8,8,−256) ( - 8,8 , - 256 ) (−8,8,−256)  
D)    (−8,−4,0) ( - 8 , - 4,0 ) (−8,−4,0)  
E)    (8,8,−256) ( 8,8 , - 256 ) (8,8,−256)

Question 6

Your company manufactures two models of stereo speakers, the Ultra Mini and the Big Stack. Demand for each depends partly on the price of the other. If one is expensive then more people will buy the other. If p1 is the price per pair of the Ultra Mini and p2 is the price of the Big Stack, demand for the Ultra Mini is given by    q1(p1,p2) =50,000−100p1+10p2q _ { 1 } \left( p _ { 1 } , p _ { 2 } \right)  = 50,000 - 100 p _ { 1 } + 10 p _ { 2 }q1​(p1​,p2​) =50,000−100p1​+10p2​   
Where q1 represents the number of pairs of Ultra Minis that will be sold in a year. The demand for the Big Stack is given by
   q2(p1,p2) =150,000+10p1−100p2q _ { 2 } \left( p _ { 1 } , p _ { 2 } \right)  = 150,000 + 10 p _ { 1 } - 100 p _ { 2 }q2​(p1​,p2​) =150,000+10p1​−100p2​   
Find the prices for the Ultra Mini and the Big Stack that will maximize your total revenue.
 
Round your answer to the nearest dollar.

Accepted Answer

A)  $798 for Ultra Mini and $275 for Big Stack
B)   $428 for Ultra Mini and $783 for Big Stack
C)   $275 for Ultra Mini and $798 for Big Stack
D)   $328 for Ultra Mini and $783 for Big Stack
E)   $783 for Ultra Mini and $328 for Big Stack

Question 7

Let $H = f _ { x x } ( a , b ) f _ { y y } ( a , b ) - f ^ { 2 } x y ( a , b )$  
What condition on H guarantees that f has a relative extremum at the point $( a , b )$

Accepted Answer

The answer of Let \(H = f _ { x...

Question 8

How many critical points does the function have     f(x,y) =2xeyf ( x , y )  = 2 x e ^ { y }f(x,y) =2xey

Accepted Answer

A)  three
B)   two
C)   four
D)   one
E)   none

Question 9

What point on the surface z is closest to the origin 
  $z = x ^ { 2 } + y - 2$  
Hint : Minimize the square of the distance from $( x , y , z )$ to the origin.
 
NOTE: Please enter your answer(s) in the form $( x , y , z )$ . If there is more than one answer, separate them with commas.

Accepted Answer

The answer of What point on the surface z is...

Question 10

Compute the integral.    ∫04∫08ex+y dx dy\int _ { 0 } ^ { 4 } \int _ { 0 } ^ { 8 } e ^ { x + y } \mathrm {~d} x \mathrm {~d} y∫04​∫08​ex+y dx dy

Accepted Answer

A)    e8e4e ^ { 8 } e ^ { 4 }e8e4 
B)    (e8−1) (e4−1) \left( e ^ { 8 } - 1 \right)  \left( e ^ { 4 } - 1 \right) (e8−1) (e4−1)  
C)    e12+12e ^ { 12 } + 12e12+12 
D)    e12−12e ^ { 12 } - 12e12−12 
E)    e4(e5−1) \frac { e ^ { 4 } } { \left( e ^ { 5 } - 1 \right)  }(e5−1) e4​

Question 11

For the function    p(x,y) =x−0.2y+0.44xyp ( x , y )  = x - 0.2 y + 0.44 x yp(x,y) =x−0.2y+0.44xy   
Find  p(40,20) p ( 40,20 ) p(40,20)   .

Accepted Answer

A)  364
B)   396
C)   388
D)   376
E)   752

Question 12

Solve the given problem by using substitution.  
Minimize  S=xy+25xz+5yzS = x y + 25 x z + 5 y zS=xy+25xz+5yz  subject to  xyz=1x y z = 1xyz=1  with   0">x>0x > 0x>0  ,   0">y>0y > 0y>0  ,   0">z>0z > 0z>0  .

Accepted Answer

A)    Smin⁡=20S _ { \min } = 20Smin​=20 
B)    Smin⁡=15S _ { \min } = 15Smin​=15 
C)    Smin⁡=5S _ { \min } = 5Smin​=5 
D)    Smin⁡=25S _ { \min } = 25Smin​=25 
E)    Smin⁡=1S _ { \min } = 1Smin​=1

Question 13

Solve the given problem by using substitution.  
Find the maximum value of  f(x,y,z) =8−x2−y2−z2f ( x , y , z )  = 8 - x ^ { 2 } - y ^ { 2 } - z ^ { 2 }f(x,y,z) =8−x2−y2−z2  subject to  z=3yz = 3 yz=3y  .

Accepted Answer

A)    fmax⁡=8f _ { \max } = 8fmax​=8 
B)    fmax⁡=9f _ { \max } = 9fmax​=9 
C)    fmax⁡=3f _ { \max } = 3fmax​=3 
D)    fmax⁡=10f _ { \max } = 10fmax​=10 
E)    fmax⁡=7f _ { \max } = 7fmax​=7

Question 14

Find  ∬Rf(x,y) dx dy\iint _ { R } f ( x , y )  \mathrm { d } x \mathrm {~d} y∬R​f(x,y) dx dy  , where  f(x,y) =xy2f ( x , y )  = x y ^ { 2 }f(x,y) =xy2  and R is the indicated domain. (Remember that you often have a choice as to the order of integration.)        f(y) =9−y2f ( y )  = \sqrt { 9 - y ^ { 2 } }f(y) =9−y2​

Accepted Answer

A)  32.4
B)   24.3
C)   16.2
D)   27
E)   259.2

Question 15

Locate the saddle point of the function.    f(x,y) =x2−6x+y−eyf ( x , y )  = x ^ { 2 } - 6 x + y - e ^ { y }f(x,y) =x2−6x+y−ey

Accepted Answer

A)    (6,1,−10) ( 6,1 , - 10 ) (6,1,−10)  
B)    (6,1,−1) ( 6,1 , - 1 ) (6,1,−1)  
C)    (3,0,−10) ( 3,0 , - 10 ) (3,0,−10)  
D)    (6,0,−10) ( 6,0 , - 10 ) (6,0,−10)  
E)    (3,1,−10) ( 3,1 , - 10 ) (3,1,−10)

Question 16

Compute the integral.    ∫010∫4−x25−x(x+y) 12 dy dx\int _ { 0 } ^ { 10 } \int _ { 4 - x } ^ { 25 - x } ( x + y )  ^ { \frac { 1 } { 2 } } \mathrm {~d} y \mathrm {~d} x∫010​∫4−x25−x​(x+y) 21​ dy dx

Accepted Answer

A)  60
B)   15
C)   -52
D)   780
E)   117

Question 17

Write the integral with the order of integration reversed (changing the limits of integration as necessary) .    ∫−11∫01+yf(x,y) dx dy\int _ { - 1 } ^ { 1 } \int _ { 0 } ^ { \sqrt { 1 + y } } f ( x , y )  \mathrm { d } x \mathrm {~d} y∫−11​∫01+y​​f(x,y) dx dy

Accepted Answer

A)    ∫02∫x−11f(x,y) dy dx\int _ { 0 } ^ { \sqrt { 2 } } \int _ { \sqrt { x } - 1 } ^ { 1 } f ( x , y )  \mathrm { d } y \mathrm {~d} x∫02c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​∫xc-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​−11​f(x,y) dy dx 
B)    ∫02∫x2−11f(x,y) dy dx\int _ { 0 } ^ { \sqrt { 2 } } \int _ { x ^ { 2 } - 1 } ^ { 1 } f ( x , y )  \mathrm { d } y \mathrm {~d} x∫02c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​∫x2−11​f(x,y) dy dx 
C)    ∫01∫1x2−1f(x,y) dy dx\int _ { 0 } ^ { 1 } \int _ { 1 } ^ { x ^ { 2 } - 1 } f ( x , y )  \mathrm { d } y \mathrm {~d} x∫01​∫1x2−1​f(x,y) dy dx 
D)      ∫−10∫x2−11f(x,y) dy dx\int _ { - 1 } ^ { 0 } \int _ { x ^ { 2 } - 1 } ^ { 1 } f ( x , y )  \mathrm { d } y \mathrm {~d} x∫−10​∫x2−11​f(x,y) dy dx 
E)      ∫02∫x−11f(x,y) dy dx\int _ { 0 } ^ { \sqrt { 2 } } \int _ { \sqrt { x - 1 } } ^ { 1 } f ( x , y )  \mathrm { d } y \mathrm {~d} x∫02c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​​∫x−1c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z'/>​1​f(x,y) dy dx

Question 18

Locate the local maximum of the function.    f(x,y) =e−(x6+y2) f ( x , y )  = e ^ { - \left( x ^ { 6 } + y ^ { 2 } \right)  }f(x,y) =e−(x6+y2)

Accepted Answer

A)    (0,0,1) ( 0,0,1 ) (0,0,1)  
B)    (1,0,1e) \left( 1,0 , \frac { 1 } { e } \right) (1,0,e1​)  
C)    (0,0,0) ( 0,0,0 ) (0,0,0)  
D)    (1,0,e) ( 1,0 , e ) (1,0,e)  
E)    (1,0,1e2) \left( 1,0 , \frac { 1 } { e ^ { 2 } } \right) (1,0,e21​)

Question 19

Use Lagrange Multipliers to solve the problem.  
Find the maximum value of  f(x,y) =3xyf ( x , y )  = 3 x yf(x,y) =3xy  subject to  x2+y2=8x ^ { 2 } + y ^ { 2 } = 8x2+y2=8  .

Accepted Answer

A)    fmax⁡=24f _ { \max } = 24fmax​=24 
B)    fmax⁡=12f _ { \max } = 12fmax​=12 
C)    fmax⁡=17f _ { \max } = 17fmax​=17 
D)    fmax⁡=7f _ { \max } = 7fmax​=7 
E)    fmax⁡=32f _ { \max } = 32fmax​=32

Question 20

Compute the integral.    ∫09∫−x2x2x dy dx\int _ { 0 } ^ { 9 } \int _ { - x ^ { 2 } } ^ { x ^ { 2 } } x \mathrm {~d} y \mathrm {~d} x∫09​∫−x2x2​x dy dx

Accepted Answer

A)  3,280.5
B)   9
C)   6,561
D)   -40.5
E)   0

Exam 15: Functions of Several Variables

Solve the given problem by using substitution. Find the maximum value of $f ( x , y , z ) = 8 - x ^ { 2 } - y ^ { 2 } - z ^ { 2 }$ subject to $z = 3 y$ .

Correct Answer
verified

Locate the local maximum of the function. $f ( x , y ) = x ^ { 2 } y - 8 x ^ { 2 } - 4 y ^ { 2 }$

Correct Answer
verified

Compute the integral. $\int _ { 0 } ^ { 4 } \int _ { 0 } ^ { 8 } e ^ { x + y } \mathrm {~d} x \mathrm {~d} y$

Correct Answer
verified

Correct Answer
verified

What point on the surface z is closest to the origin $z = x ^ { 2 } + y - 2$ Hint : Minimize the square of the distance from $( x , y , z )$ to the origin. NOTE: Please enter your answer(s) in the form $( x , y , z )$ . If there is more than one answer, separate them with commas.

Correct Answer
verified

How many critical points does the function have $f ( x , y ) = 2 x e ^ { y }$

Correct Answer
verified

Correct Answer
verified

Solve the given problem by using substitution. Minimize $S = x y + 25 x z + 5 y z$ subject to $x y z = 1$ with $x > 0$ , $y > 0$ , $z > 0$ .

Correct Answer
verified

Use Lagrange Multipliers to solve the problem. Find the maximum value of $f ( x , y ) = 3 x y$ subject to $x ^ { 2 } + y ^ { 2 } = 8$ .

Correct Answer
verified

Compute the integral. $\int _ { 0 } ^ { 10 } \int _ { 4 - x } ^ { 25 - x } ( x + y ) ^ { \frac { 1 } { 2 } } \mathrm {~d} y \mathrm {~d} x$

Correct Answer
verified

Let $H = f _ { x x } ( a , b ) f _ { y y } ( a , b ) - f ^ { 2 } x y ( a , b )$ What condition on H guarantees that f has a relative extremum at the point $( a , b )$

Correct Answer
verified

Correct Answer
verified

Evaluate $f ( a , 2 )$ for $f ( x , y ) = x ^ { 2 } - y - x y + 5$

Correct Answer
verified

Locate the saddle point of the function. $f ( x , y ) = x ^ { 2 } - 6 x + y - e ^ { y }$

Correct Answer
verified

For the function $p ( x , y ) = x - 0.2 y + 0.44 x y$ Find $p ( 40,20 )$ .

Correct Answer
verified

Locate the local maximum of the function. $f ( x , y ) = e ^ { - \left( x ^ { 6 } + y ^ { 2 } \right) }$

Correct Answer
verified

Compute the integral. $\int _ { 0 } ^ { 9 } \int _ { - x ^ { 2 } } ^ { x ^ { 2 } } x \mathrm {~d} y \mathrm {~d} x$

Correct Answer
verified

Correct Answer
verified

Write the integral with the order of integration reversed (changing the limits of integration as necessary) . $\int _ { - 1 } ^ { 1 } \int _ { 0 } ^ { \sqrt { 1 + y } } f ( x , y ) \mathrm { d } x \mathrm {~d} y$

Correct Answer
verified

Use Lagrange Multipliers to solve the given problem. Find the maximum value of $f ( x , y ) = x y$ subject to $2 x + y = 60$ .

Correct Answer
verified

Exam 15: Functions of Several Variables

Solve the given problem by using substitution. Find the maximum value of f(x,y,z) =8−x2−y2−z2f ( x , y , z ) = 8 - x ^ { 2 } - y ^ { 2 } - z ^ { 2 }f(x,y,z) =8−x2−y2−z2 subject to z=3yz = 3 yz=3y .

Correct AnswerverifiedShow Answer

Locate the local maximum of the function. f(x,y) =x2y−8x2−4y2f ( x , y ) = x ^ { 2 } y - 8 x ^ { 2 } - 4 y ^ { 2 }f(x,y) =x2y−8x2−4y2

Correct AnswerverifiedShow Answer

Compute the integral. ∫04∫08ex+y dx dy\int _ { 0 } ^ { 4 } \int _ { 0 } ^ { 8 } e ^ { x + y } \mathrm {~d} x \mathrm {~d} y∫04​∫08​ex+y dx dy

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

How many critical points does the function have f(x,y) =2xeyf ( x , y ) = 2 x e ^ { y }f(x,y) =2xey

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Solve the given problem by using substitution. Minimize S=xy+25xz+5yzS = x y + 25 x z + 5 y zS=xy+25xz+5yz subject to xyz=1x y z = 1xyz=1 with x>0x > 0x>0 , y>0y > 0y>0 , z>0z > 0z>0 .

Correct AnswerverifiedShow Answer

Use Lagrange Multipliers to solve the problem. Find the maximum value of f(x,y) =3xyf ( x , y ) = 3 x yf(x,y) =3xy subject to x2+y2=8x ^ { 2 } + y ^ { 2 } = 8x2+y2=8 .

Correct AnswerverifiedShow Answer

Compute the integral. ∫010∫4−x25−x(x+y) 12 dy dx\int _ { 0 } ^ { 10 } \int _ { 4 - x } ^ { 25 - x } ( x + y ) ^ { \frac { 1 } { 2 } } \mathrm {~d} y \mathrm {~d} x∫010​∫4−x25−x​(x+y) 21​ dy dx

Correct AnswerverifiedShow Answer

Let H=fxx(a,b)fyy(a,b)−f2xy(a,b)H = f _ { x x } ( a , b ) f _ { y y } ( a , b ) - f ^ { 2 } x y ( a , b )H=fxx​(a,b)fyy​(a,b)−f2xy(a,b) What condition on H guarantees that f has a relative extremum at the point (a,b)( a , b )(a,b)

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Evaluate f(a,2) f ( a , 2 ) f(a,2) for f(x,y) =x2−y−xy+5f ( x , y ) = x ^ { 2 } - y - x y + 5f(x,y) =x2−y−xy+5

Correct AnswerverifiedShow Answer

Locate the saddle point of the function. f(x,y) =x2−6x+y−eyf ( x , y ) = x ^ { 2 } - 6 x + y - e ^ { y }f(x,y) =x2−6x+y−ey

Correct AnswerverifiedShow Answer

For the function p(x,y) =x−0.2y+0.44xyp ( x , y ) = x - 0.2 y + 0.44 x yp(x,y) =x−0.2y+0.44xy Find p(40,20) p ( 40,20 ) p(40,20) .

Correct AnswerverifiedShow Answer

Locate the local maximum of the function. f(x,y) =e−(x6+y2) f ( x , y ) = e ^ { - \left( x ^ { 6 } + y ^ { 2 } \right) }f(x,y) =e−(x6+y2)

Correct AnswerverifiedShow Answer

Compute the integral. ∫09∫−x2x2x dy dx\int _ { 0 } ^ { 9 } \int _ { - x ^ { 2 } } ^ { x ^ { 2 } } x \mathrm {~d} y \mathrm {~d} x∫09​∫−x2x2​x dy dx

Correct AnswerverifiedShow Answer

Correct AnswerverifiedShow Answer

Write the integral with the order of integration reversed (changing the limits of integration as necessary) . ∫−11∫01+yf(x,y) dx dy\int _ { - 1 } ^ { 1 } \int _ { 0 } ^ { \sqrt { 1 + y } } f ( x , y ) \mathrm { d } x \mathrm {~d} y∫−11​∫01+y​​f(x,y) dx dy

Correct AnswerverifiedShow Answer

Use Lagrange Multipliers to solve the given problem. Find the maximum value of f(x,y) =xyf ( x , y ) = x yf(x,y) =xy subject to 2x+y=602 x + y = 602x+y=60 .

Correct AnswerverifiedShow Answer

Solve the given problem by using substitution. Find the maximum value of $f ( x , y , z ) = 8 - x ^ { 2 } - y ^ { 2 } - z ^ { 2 }$ subject to $z = 3 y$ .

Correct Answer
verified

Locate the local maximum of the function. $f ( x , y ) = x ^ { 2 } y - 8 x ^ { 2 } - 4 y ^ { 2 }$

Correct Answer
verified

Compute the integral. $\int _ { 0 } ^ { 4 } \int _ { 0 } ^ { 8 } e ^ { x + y } \mathrm {~d} x \mathrm {~d} y$

Correct Answer
verified

Correct Answer
verified

Correct Answer
verified

How many critical points does the function have $f ( x , y ) = 2 x e ^ { y }$

Correct Answer
verified

Correct Answer
verified

Solve the given problem by using substitution. Minimize $S = x y + 25 x z + 5 y z$ subject to $x y z = 1$ with $x > 0$ , $y > 0$ , $z > 0$ .

Correct Answer
verified

Use Lagrange Multipliers to solve the problem. Find the maximum value of $f ( x , y ) = 3 x y$ subject to $x ^ { 2 } + y ^ { 2 } = 8$ .

Correct Answer
verified

Compute the integral. $\int _ { 0 } ^ { 10 } \int _ { 4 - x } ^ { 25 - x } ( x + y ) ^ { \frac { 1 } { 2 } } \mathrm {~d} y \mathrm {~d} x$

Correct Answer
verified

Let $H = f _ { x x } ( a , b ) f _ { y y } ( a , b ) - f ^ { 2 } x y ( a , b )$ What condition on H guarantees that f has a relative extremum at the point $( a , b )$

Correct Answer
verified

Correct Answer
verified

Evaluate $f ( a , 2 )$ for $f ( x , y ) = x ^ { 2 } - y - x y + 5$

Correct Answer
verified

Locate the saddle point of the function. $f ( x , y ) = x ^ { 2 } - 6 x + y - e ^ { y }$

Correct Answer
verified

For the function $p ( x , y ) = x - 0.2 y + 0.44 x y$ Find $p ( 40,20 )$ .

Correct Answer
verified

Locate the local maximum of the function. $f ( x , y ) = e ^ { - \left( x ^ { 6 } + y ^ { 2 } \right) }$

Correct Answer
verified

Compute the integral. $\int _ { 0 } ^ { 9 } \int _ { - x ^ { 2 } } ^ { x ^ { 2 } } x \mathrm {~d} y \mathrm {~d} x$

Correct Answer
verified

Correct Answer
verified

Write the integral with the order of integration reversed (changing the limits of integration as necessary) . $\int _ { - 1 } ^ { 1 } \int _ { 0 } ^ { \sqrt { 1 + y } } f ( x , y ) \mathrm { d } x \mathrm {~d} y$

Correct Answer
verified

Use Lagrange Multipliers to solve the given problem. Find the maximum value of $f ( x , y ) = x y$ subject to $2 x + y = 60$ .

Correct Answer
verified