Math B6C – Chapter 15 Quiz – Spring 2007

* Solutions* __________________________________________________________________________________

1. Find the limit as x approaches 0 along the x-axis: ( ) ( )

8 8

8 8,0 0,0

15 4lim ?5 2x

x yx y→

− =+

( ) ( )

8 8 8 8

8 8 8 80 0,0 0,0315 4 15 4 0 15lim lim lim lim

5 2 5 2 0 5x xx

x y x xx y x x→ →→

− − ⋅= =+ + ⋅ 0x→

=

( ) ( )

8 8

8 8,0 0,0

15 4lim 35 2x

x yx y→

− =+

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2. Find the limit as y approaches 0 along the y-axis: ( ) ( )

8 8

8 80, 0,0

15 4lim ?5 2y

x yx y→

− =+

( ) ( )

8 8 8 8

8 8 8 80 00, 0,0

15 4 15 0 4 4 4lim lim lim lim5 2 5 0 2 2 2y yy

x y y yx y y y→ →→

− ⋅ − −= = =+ ⋅ + 0y→

−

( ) ( )

8 8

8 80, 0,0

15 4lim 25 2y

x yx y→

− = −+

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3. Find the limit: ( ) ( )

8 8

8 8, 0,0

15 4lim ?5 2x y

x yx y→

− =+

( ) ( )

(by problem #18 8

8 8,0 0,0 15 4lim 3

5 2x

x yx y→

− =+

) , but

( ) ( )

( )by problem #28 8

8 80, 0,0 15 4lim 2

5 2y

x yx y→

− = −+

.

Since these two limits are unequal, ( ) ( )

8 8

8, 0,0

15 4lim5 2x y 8

x yx y→

−+

does not exist.

__________________________________________________________________________________ 4. If ( ) ( )( )3, ln cosy yg x y x π= + , then ( ) ?2,2xg =

( ) ( )( )( )( )

( )( )( )

( ) ( )( )

3 23

3 3

2

3

cos 3 cos, ln cos

cos cos3 cos

,cos

x

x

y yy y y

y y

y x y

yx yxg x y x

x xx y

g x yx

πxπ π

π

π π

π

∂ +∂ ∂= + = =∂ +

=+

+

Hence,

( ) ( )( )

2

32 2 23 2 cos 2 3 4 3 4 62,2

2 cos 8 10 5xgππ

⋅⋅ ⋅ ⋅ ⋅= = =+ +

= .

__________________________________________________________________________________

5. If ( ) ( )cos, x yh x y e −= , then ( ) ?1,1xyh =

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

cos

cos cos

cos cos cos

cos cos

cos sin

sin sin sin

sin

, , , x yxy x

x y x y

x y x y x y

x y x

y y x y x

x y x y x yy x y x

x y x yy y

x y x yy

h x y h x y h x y e

e e

e e e

e e

−

− −

− − −

−

∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞− − −⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

⎛ ⎞∂ ∂ x yy∂

− − − −⎜ ⎟∂ ∂ ∂⎝ ⎠⎛ ⎞∂

− − −⎜ ⎟∂⎝ ⎠

= = =

= = −

= − = −

= − ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )cos cos

cos

sin cos,

y

x y x yxy

x y x yy

x y x yh x y e e

−

− −

∂− −

∂

− + −=

Hence, ( ) ( ) ( ) ( ) ( )cos 1 1 cos 1 1 cos0 cos0 1sin 1 1 cos 1 1 0 cos0 0 11,1xyh e e e e e− −− + − ⋅ + + ⋅= = = , so

( )1,1xyh e=

__________________________________________________________________________________ 6. Find the linearization of ( ) 5 2, 5 2f x y x y x y= − + at ( )1, 2 . ( ) ?,L x y =

( ) ( )

( ) [ ]

( ) ( ) [ ]

( ) ( )

5 2 5 2

4 2 5

'( , ) 5 2 5 2

'( , ) 5 5 2 2

' 1, 2 15 6

1 1( , ) 1, 2 ' 1, 2 3 15 6

2 2

( , ) 3 15 1 6 2 3 15 15 6 12

x yf x y f f x y x y x y x yx y

f x y x y x y

f

x xL x y f f

y y

L x y x y x y

⎡ ⎤∂ ∂⎡ ⎤= = − + − +⎢ ⎥⎣ ⎦ ∂ ∂⎣ ⎦

⎡ ⎤= − +⎣ ⎦

=

−⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + − = +⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠

= + − + − = + − + −

( , ) 15 6 24L x y x y= + − A graph of ( ) 5 2, 5 2f x y x y x y= − + along with its linearization at ( )1, 2 , : ( , ) 15 6 24L x y x y= + −

__________________________________________________________________________________

7. Find the linearization of ( ) 3 2 4, , sinh( )f x y z x z y xy= − at ( )1,0,2 . ( ) ?, ,L x y z =

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )

( ) [ ]

3 2 4 3 2 4 3 2 4

2 2 4 3 4 3

2 2 5 3 4 3

'( , , )

sinh( ) sinh( ) sinh( )

3 cosh 4 sinh cosh 2

'( , , ) 3 cosh 4 sinh cosh 2

' 1,0,2 12 0 4

( , , ) 1

x y zf x y z f f f

x z y xy x z y xy x z y xyx y z

x z y xy y y xy y xy x x z

f x y z x z y xy y xy xy xy x z

f

L x y z f

⎡ ⎤= ⎣ ⎦⎡ ⎤∂ ∂ ∂

= − − −⎢ ⎥∂ ∂ ∂⎣ ⎦⎡ ⎤= − ⋅ + ⋅⎣ ⎦⎡ ⎤= − +⎣ ⎦

=

= ( ) ( ) [ ]

( ) ( )

1 1,0, 2 ' 1,0, 2 0 4 12 0 4

2 2

( , , ) 4 12 1 0 4 2

x xf y y

z z

L x y z x z

⎛ ⎞ −⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ − = +⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠

= + − + + −

( , , ) 12 4 16L x y z x z= + −

The level surface

3 2 4 sinh( ) 4x z y xy− = :

__________________________________________________________________________________

8. Find the gradient of ( ) ( )3 3 2 2cos, x y x yg x y = at the point ( . )2,1 Rounded off to the nearest thousandth, what is the x-component of this gradient?

( )3 3 2 2cosz x y x y=

( ) ( ) ( ) ( )( ) ( )( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( )

3 3 2 2 3 3 2 2

3 3 2 2 3 3 2 2 3 3 2 2 3 3 2 2

2 3 2 2 3 3 2 2 2 3 2 2 2 3 3 2 2 2

2 3 2 2 4 5

, , , , cos , cos

cos cos , cos cos

3 cos sin 2 , 3 cos sin 2

3 cos 2 si

x yg x y g x y g x y x y x y x y x yx y

x y x y x y x y x y x y x y x yx x y y

x y x y x y x y xy x y x y x y x y x y

x y x y x y

∂ ∂∇ = =

∂ ∂

⎛ ⎞∂ ∂ ∂ ∂⎛ ⎞= + +⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

= − ⋅ − ⋅

= − ( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

2 2 3 2 2 2 5 4 2 2

2 3 2 2 4 5 2 2 3 2 2 2 5 4 2 2

n , 3 cos 2 sin

2,1 3 2 1 cos 2 1 2 2 1 sin 2 1 , 3 2 1 cos 2 1 2 2 1 sin 2 1

12,1 12cos 4 32sin 4 , 24cos 4 64sin 4 12cos 4 32sin 4

2

x y x y x y x y x y

g

g

−

∇ = ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅

⎡ ⎤∇ = − − = − ⎢ ⎥

⎣ ⎦

The x-component of this gradient is thus ( ) ( )12cos 4 32sin 4 16.373956399− ≅ , about 16.374 . __________________________________________________________________________________

9. Find the gradient of ( ) 4 4 72, , y zf x y z x −= at the point ( )1,1,1 . Rounded off to the nearest thousandth, the z-component of this gradient is _____ .

( ) ( ) ( ) ( )

( )

( )

4 4 7 4 4 7 4 4 7

3 4 4 3 6

, , 2 , 2 , 2

, , 4 , 4 , 14

1,1,1 4, 4, 14

f x y z y z y z y zx y z

f x y z x y x y z

f

x x x∂ ∂ ∂∇ = − − −

∂ ∂ ∂

∇ = −

∇ = −

The z-component of this gradient is thus 14− .

The graph depicts the gradient field,

( ) 3 4 4 3 6, , 4 , 4 , 14f x y z x y x y z∇ = −

which is always perpendicular to f ’s level surfaces, one of which is also shown:

( ) 4 4 72 1, , y zf x y z x − = −= __________________________________________________________________________________

10. Find , for the f featured in problem #9, with (ˆ 1,1,1uD f )1

1ˆ 214

3u

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

.

( ) ( )

( )

ˆ

4 1 2 11 14 2 2 2 214 1414 3 7 3

ˆ1,1,1 1,1,1

2 12 2 30 30 142 2 2 4 21 14

1 1414 14 147 3 4

uD f f u⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= ∇

− −= = + − = =−

i ii

i

( )ˆ15 141,1,1

7uD f −=

__________________________________________________________________________________

11. In which direction is the temperature function ( ) 2arctan, , xzy

T x y z ⎛⎜⎝ ⎠

= ⎞⎟

)

increasing most rapidly, from

the point ( ? Find a vector in this desired direction, and with a magnitude equal to the rate of 1, 2, 4 increase in this direction. The gradient points in the direction in which the scalar temperature field T is increasing most rapidly, and also has the desired magnitude. So we’ll calculate this gradient, and evaluate it at the point .

T∇( )1, 2, 4

( ) ( ) ( ) ( )

( ) ( ) ( )

( )( )

( )( )

( )( )

2 2 2

22 2

2 2 22 2 2

2 3 2

2 2 4 2 2 4 2 2 4

2 2 4 2 3 2

, , , , , , , , , ,

arctan , arctan , arctan

, ,1 1 1

2, ,1 1 1

1 2 1, ,1

x y zT x y z T x y z T x y z T x y z

xzy xzy xzyx y z

xzyxzy xzyyx zxzy xzy xzy

zy xzy xyx z y x z y x z y

z xz xx z y y y y y

− − −

−− −

− − −

− − −

− − −

−

∇ =

∂ ∂ ∂=

∂ ∂ ∂

∂∂ ∂∂∂ ∂=

+ + +

−=

+ + +

−= =

+

( )

( )

2 2 2 2

2

4 2 2

2

4 2 2 2 2

2, ,

2, , , ,

2 2 1 4 1 11, 2, 4 4, , 1 4, 4, 1 4, 4, 12 1 4 2 2 1 4 8

xzz xx z y y

y xzT x y z z xy x z y

T

−

−+

−∇ =

+

− ⋅ ⋅∇ = = − = −

+ ⋅ + ⋅

( ) 1 1 11,2,4 , ,2 2 8

T∇ = −

The gradient field,

( )2

4 2 2

2, , , ,y xzT x y z z xy x z y

−∇ =

+

__________________________________________________________________________________

12. How fast is the temperature function ( ) 2arctan, , xzy

T x y z ⎛⎜⎝ ⎠

= ⎞⎟ increasing at the point ,

in its direction of maximum increase? Round off your answer to the nearest hundred-thousandth.

( )1, 2, 4

The gradient points in the direction in which the temperature T is increasing most rapidly, and its magnitude is the rate of increase in this direction. We calculated this gradient in problem #11, so now we need but find this vector’s length:

T∇

( )2 2 21 1 1 1 1 1 1 1 1 331,2,4 , ,

2 2 8 2 2 8 4 4 64 8T ⎛ ⎞ ⎛ ⎞ ⎛ ⎞∇ = − = + − + = + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( )1,2,4 0.71807T∇ ≅

NOTE: If distance is measured in centimeters and temperature measured in degrees Celsius, then the quantity we found has units of degrees Celsius per centimeter.

__________________________________________________________________________________ 13. Find the directional derivative of ( ), , yz xzW x y z xe ye= − at the point ( )3,2,0

in the direction of vector 5, 1, 1v = − . We need a unit vector in the direction of v :

( )22 2 3

51

31

5, 1,1 5, 1,1 5, 1,1ˆ27 35 1 1

1ˆ3

vuv

u⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥⎣ ⎦

− − −= = = =

+ − +

=

Next we need the gradient of W at the point ( )3,2,0 :

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) 0 0 0 0

, , , , , , , , , ,

, ,

, ,

3,2,0 0, 0 , 6 6 1, 1, 0

x y z

yz xz yz xz yz xz

yz xz yz xz yz xz

x y z

x y z

W W x y z W x y z W x y z

xe ye xe ye xe ye

e yze xze e xye yxe

W e e e e

∂ ∂ ∂=

∂ ∂ ∂

−

−

∇ =

− − −

= − −

∇ = − − = −

Now we can compute the desired directional derivative:

( ) ( )

( )

ˆ 3, 2,0 3, 2,0

2 35 1 03

ˆ1 5 1 5

1 11 1 1 13 3 3 30 1 0 1

1 6 6 33 33 3 3

333

uD W W u⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

+ +

=∇

= − − = − −

= = = =⋅

i

i i

The level surface ( ) 1, , yz xzW x y z xe ye == −

containing the point ( ) . 3,2,0

14. Find the equation of the tangent plane to 2 3 42 2x z xyz y− + = at the point ( )1,1,1 . Let ( ) 2 3 4, , 2x y zF x z xyz y= − +

( ), , 2x y z =

, so we seek an equation for a tangent plane to the implicitly defined surface

. Since F F∇ is normal to the level surface, it will be normal to the tangent plane as well.

( )

( )

3 3 2 2, , 4 , 4 , 6

1,1,1 3, 3, 5

F x y z xz yz xz y x z xy

F n

∇ = − − + −

∇ = =

The point of tangency, , is on both the implicit surface(1,1,1) ( ), , 2x y zF = and the tangent plane to the graph of F at that point. So the equation of the tangent plane is given by:

, , 1,1,1

3, 3, 5 , , 3, 3, 5 1,1,1

3 3 5 3 3 5

n x y z n

x y z

x y z

=

=

+ + = + +

i i

i i

3 3 5 11x y z+ + =

The implicitly defined surface,

2 3 42 2x z xyz y− + = ,

and it’s tangent plane at the point ( ) , 1,1,1

3 3 5 11x y z+ + = :

__________________________________________________________________________________ 15. Find the equation of the tangent plane to the implicitly defined surface

( )cos sin cos 0xyz x z− − = at the point ( ),1,0π .

Let ( ) ( ), , cos sin cosx y z x zF xyz= − −

(, so we seek an equation for a tangent plane to the implicitly defined

surface ), , 0x y zF = . Since F∇ is normal to the level surface, it will be normal to the tangent plane as well.

( ) ( ) ( ) ( )

( )

, , sin cos , sin , sin sin

,1,0 0 cos , 0, 0 1, 0, 0

F x y z yz xyz x xz xyz xy xyz z

F nπ π

∇ = − − − − +

∇ = − = =

The point of tangency, , is on both the implicit surface( ,1,0π ) ( ), , 0x y zF = and the tangent plane to the graph of F at that point. So the equation of the tangent plane is given by:

,1,0, ,1, 0, 0 , , 1, 0, 0 ,1,0

n x y z nx y z

π

π

=

=

i ii i

x π=

The implicitly defined surface,

( )cos sin cos 0x zxyz − − = ,

and it’s tangent plane at the point ( ) , ,1,0π

x π=

Be grateful I didn’t ask you to graph it !

__________________________________________________________________________________ 16. Find any local extrema and saddle points of 2 2( , ) 5 2 3 4f x y x xy y x y= − + + − . Extrema and saddle points occur at any points where '( , ) 0f x y = :

'( , ) 10 3 4 4 0 0f x y x y x y⎡ ⎤⎣ ⎦= − + − + − = ⎡ ⎤⎣ ⎦

10 1 31 4 4

xy

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

− −=

−

Which has solution 8 ,39 39

x y= − = 37 . So a possible extreme value or saddle point is 8 37,39 39

⎛ ⎞⎜ ⎟⎝ ⎠− . To see

which it is, we invoke the multivariable second derivative test. We need the second derivative matrix, the Hessian:

10 11 4

xx xy

yx yy

f ff f

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦

−=

−

The determinant of this Hessian matrix is 40 – 2 = 38 > 0, and 10 0xxf = > , hence our point 8 37,39 39

⎛ ⎞⎜ ⎟⎝ ⎠− is a

relative minimum.

The surface

2 25 2 3z x xy y x= − + + − 4y ,

along with its relative minimum at the point

8 37 86, ,39 39 39

⎛ ⎞⎜ ⎟⎝ ⎠− − :

__________________________________________________________________________________

17. Find the absolute maximum and the absolute minimum of the function 3 2 2( , ) y 3f x y x x xy y−= + − over the unit disk 2 2 1x y ≤+ . First we’ll look for any critical points of f within the disk:

2 2

2 2

23 0

2 3 0x

y

yf x x y

f x xy y

−

−

= + =

= + − =

Adding these equations eliminates that pesky middle term, and allows us to solve this system of simultaneous nonlinear equations. We get:

( )( ) ( )

2 2

2 2

2 22 0

2 0

x yx y

x y x y+ −

0− =

− =

=

This equation has for it’s solutions ALL points on the lines y x= − and y x= , so the “X” in the unit disk on these lines consists entirely of critical points. We have an infinite number of critical points! ☺ First lets look at the values of f on the line y x= :

( )

3 2 2 3

3 2 2 3 3 3 3 3, 0( , ) y

x x xx xf x y x x xy yf x x x x x x x

−

− − +

= + −= + − = − =

So for all the input points on the line y x= , f outputs 0. Next consider the values of f on the line y x= − :

( ) ( ) ( ) ( )

3 2 2 3

2 33 2 3 3 3 3, 4

( , )

x x x x

y

x

f x y x x xy y3f x x x x x x x x− −

−

− +− =−

= + −

= + − = + +

So for the input point on the line ( , )a a− y x= − , f outputs . Thus along the line 34a y x= − , f attains its

minimum at the point 2 22 2

,⎛ ⎞⎜⎝ ⎠− ⎟ on the unit circle, where

324 4

2⎞

= =⎟⎠

2 2 2 22 2 8

,⎛ ⎞ ⎛ ⎛ ⎞

= −⎜ ⎟ ⎜ ⎜ ⎟⎝ ⎠ ⎝ ⎝ ⎠− − − 2f .

And along the line y x= − , f attains its maximum at the point 2 22 2

,⎛ ⎞⎜ ⎟⎝ ⎠

− , where

32 2 2 2 24 4

2 2 2 8,f

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠2= .

We still must check for possible extrema on the circular boundary of our disk. This boundary is parametrized by:

cos( )

sint x

r tt y

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Evaluating f on this circle gives us a real valued function of t , whose extrema we seek.

Letting and cosc = t sins t= :

( ) ( )

( ) ( ) ( )( ) ( )

3 2 2 3

3 2 2 3

2 2 2 2

( , )( ) ( ) ,

1

( )

3 7'( ) 0 sin cos4 4

f x y x x y xy yg t f r t f c s

c c s cs s

c c s s c s c s c s c s

g t c s

g t s c t t t π π

= − + −

= =

= − + −

= − + − = + − = ⋅ −

= −

= − − = ⇒ = − ⇒ = or

These values of t yield the points 2 2,2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

and 2 2,2 2

⎛ ⎞−⎜

⎝ ⎠⎟ , which we already found earlier.

To summarize, the possible absolute maximum (and minimum) must be one of the following:

( ) 0,

2 2,2 2

2 2

2

2,2 2

f a a

f

f

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

=

− =

←=

−

−

← absolute minimum

absolute maximum

The absolute minimum of f over the disk is 2− ,

and the absolute maximum is 2 .

3 2 2( , ) y 3f x y x x xy y−= + − __________________________________________________________________________________

Problems #18 - 21 refer to the level curve of a function, ( , ) 3f x y = , depicted in the following graph (along with some other level curves of this function:

18. At what point is it true that ( , )f x y∇ has a negative x-component and a small negative y-component? A 19. At what point is it true that ( , )f x y∇ has a negative x-component and a positive (though small) y-component? D 20. At what point is it true that ( , )f x y∇ has zero x-component and a negative y-component? B 21. At what point is it true that ( , )f x y∇ has zero x-component and a positive y-component? E

__________________________________________________________________________________

22. Suppose that and that ( , , )

( , , )( , , )

u x y zg x y z

v x y z⎡

= ⎢⎣ ⎦

⎤⎥

( , )( , )

( , )m u v

h u vn u v⎡ ⎤

= ⎢ ⎥⎣ ⎦

. Let f h g= .

Suppose further that and that 7

(0, 1,2)1

g ⎡ ⎤− = ⎢ ⎥

⎣ ⎦

( )( )

( )( )

7,1 7,1 3 17,1 7,1 1 2

u v

u v

m mn n

−⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥−⎣ ⎦⎣ ⎦

and that

(0, 1,2)(0, 1,2) (0, 1,2) 1 2 3(0, 1,2)(0, 1,2) (0, 1,2) 0 3 0

yx z

yx z

uu uvv v

−− − −⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥−− − −⎣ ⎦⎣ ⎦

. Then '(0, 1,2) ?f − =

Computing the given derivative matrices, we get:

( , , )( , , ) ( , , )'( , , )

( , , )( , , ) ( , , )yx z

yx z

u x y zu x y z u x y zg x y z

v x y zv x y z v x y z⎡ ⎤

= ⎢ ⎥⎣ ⎦

and ( , ) ( , )

'( , )( , ) ( , )

u v

u v

m u v m u vh u v

n u v n u v⎡ ⎤

= ⎢ ⎥⎣ ⎦

Applying the chain rule, we obtain:

( ) ( ) ( )

( ) ( )( )

( )( )

' , , ' ( , , ) ' , ,

( , , )( , , )( , , ) ( , , ) ( , , )' , ,

( , , )( , , )( , , ) ( , , ) ( , , )yxu v z

yxu v z

f x y z h g x y z g x y z

u x y zu x y zm g x y z m g x y z u x y zf x y z

v x y zv x y zn g x y z n g x y z v x y z

=

⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥

⎣ ⎦⎣ ⎦

Thus, ( )( )

( )( )

(0, 1,2)(0, 1,2)7,1 7,1 (0, 1,2)'(0, 1,2)

(0, 1,2)(0, 1,2)7,1 7,1 (0, 1,2)

3 1 1 2 31 2 0 3 0

3 9 9'(0, 1,2)

1 8 3

yxu v z

yxu v z

uum m uf

vvn n v

f

−− −⎡ ⎤ ⎡ ⎤− = ⎢ ⎥ ⎢ ⎥−− −⎣ ⎦⎣ ⎦

− −⎡ ⎤ ⎡ ⎤= ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦

−⎡ ⎤− = ⎢ ⎥−⎣ ⎦

__________________________________________________________________________________

23. Same situation described in problem #22. If ( , , )

( , , )( , , )x y z

f x y zx y z

ϕψ⎡ ⎤

= ⎢ ⎥⎣ ⎦

, then at the point (0, 1,2)− ,

?x y z zϕ ψ ϕ ψ∂ ∂ ∂ ∂

− =∂ ∂ ∂ ∂

In problem #22 we found that 3 9 9

'(0, 1,2)1 8 3

f−⎡ ⎤

− = ⎢ ⎥−⎣ ⎦. But this matrix is none other than the matrix:

3 9 91 8 3

yx z

x y z

δϕδϕ δϕδδ δ

δψ δψ δψδ δ δ

⎡ ⎤⎢ ⎥ −⎡ ⎤⎢ ⎥ = ⎢ ⎥−⎢ ⎥ ⎣ ⎦⎢ ⎥⎣ ⎦

Hence, ( ) ( )3 8 9 3 24 27 3x y z zϕ ψ ϕ ψ∂ ∂ ∂ ∂

− = ⋅ − − − = − = −∂ ∂ ∂ ∂

.

__________________________________________________________________________________

24. Find the maximum and minimum values of the function ( , , )f x y z yz xz xy= − + on the unit sphere . 2 2 2 1y zx + + = We’ll use Lagrange multipliers to maximize f subject to the constraint equation . 2 2 2 1( , , ) y zG x y z x= + + =

Then f Gλ∇ = ∇ at the extreme values of f on a level curve of . Calculating the gradients, and setting up the Lagrange multiplier equations, we get:

G

( ) ( ), , 2 , 2 , 2, , & , ,y z z x y x x y zf x y z G x y z− + −∇ = ∇ =

222

y z xz x yy x z

λλλ

− =+ =− =

Adding the first two equations, we get:

( ) ( ) ( )2 2 2 0 2 1x y x y x y x y xλ λ λ λ+ = + + = = −⇒ y+ +⇒

So either or y = −x 12

λ = . We’ll investigate the case where 12

λ ≠ below.

Adding the second two equations, we get:

( ) ( ) ( )2 2 2 0 2 1y z y z y z y z yλ λ λ λ+ = + + = = −⇒ z+ +⇒

So either or y = −z 12

λ = . We’ll investigate the case where 12

λ ≠ below.

In the case where 12

λ = , the original equations become:

y z xz x yy x z

− =+ =− =

Now simply notice that any point with coordinates such that x=y and z=0 (i.e., any point of the form ( ), ,0a a ), satisfies all three equations. Or if you must, you can always find a parametrization of the line of intersection of these two planes. Oops, I mean three planes, but the first two equations are equations of the same plane, so yea, I really did mean two planes. Take the cross product of the normal vectors to these distinct planes to get a direction vector for the line, which must be through the origin since the planes all pass through that common point. But after all that work, you’ll be convinced that x=y and z=0. Substituting these values into our constraint equation, we get:

2 2 2 2 2 2 11 0 1 2

y z xx x x+ + = +⎯ + =⎯→ ⇒ y= ± =

So possible extrema for f on the unit sphere are the two points

1 1, ,02 2

⎛ ⎞⎜ ⎟⎝ ⎠

, where 1 1, ,022 2

f ⎛ ⎞ 1=⎜ ⎟

⎝ ⎠,

and 1 1, ,2 2

⎛ ⎞− −⎜ ⎟⎝ ⎠

0 , where 1 1, ,022 2

f ⎛ ⎞ 1− − =⎜ ⎟⎝ ⎠

.

Now suppose 12

λ ≠ . Then we know from above that y x= − and that y z= − . But that immediately implies

that x z= . So we can express all the variables in terms of z: , , x z y z z z= = − = . Substituting these values into our constraint equation, we get:

( )22 2 2 11 3 1 3

z zz z⎯⎯→ ⇒+ − + = = z x= ± =

And since , these variables have opposite signs. We’ve found two points, y = −z 1 1 1, ,3 3 3± ±⎛ ⎞

⎜ ⎟⎝ ⎠

∓ .

So possible extrema for f on the unit sphere are the two points

1 1 1, ,3 3 3

⎛ ⎞−⎜ ⎟⎝ ⎠

,

where

1 1 1 1 1 1 1 1 1 1 1, ,3 3 33 3 3 3 3 3 3 3 3

1 1 1, , 13 3 3

f

f

⎛ ⎞ ⎛ ⎞− = − − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞− = −⎜ ⎟⎝ ⎠

1− + − −

and 1 1 1, ,3 3 3

⎛ ⎞− −⎜ ⎟⎝ ⎠

where

1 1 1 1 1 1 1 1 1 1 1 1, ,3 3 33 3 3 3 3 3 3 3 3

1 1 1, , 13 3 3

f

f

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− − = − − − − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

⎛ ⎞− − = −⎜ ⎟⎝ ⎠

− + − −

We’ve found two absolute maximums for f on the unit sphere, and two absolute minimums:

absolute maximum

minimum

1 1 1 1 1 1 1, ,0 0 0 22 2 2 2 2 2

1 1 1 1 1 1 1 1 1, absolute

, 1 3 3 3 3 3 3 3

3

3

( , , )

f

f x y z y z x z x y

f

⋅ ⋅ ⋅

⎛ ⎞ ⋅ − ⋅ + ⋅ =⎜ ⎟⎝ ⎠± ± ± ± ± ±⎛ ⎞ ⋅ − ⋅

←

+ ⋅ = ←−⎜ ⎟⎝ ⎠

= − +± ± ± ± ± ±=

=∓ ∓ ∓

__________________________________________________________________________________ 25. Given that a tetrahedron has volume = 1, what is the minimum possible value that A h+ can have, where A is the area of the tetrahedron and is its height?

h

(Hints: The area of a tetrahedron is the one third the area if its base times its height. Also, ignore units for this problem.)

We’ll use Lagrange multipliers to minimize the quantity volume A h+ subject to the constraint equation

113

Ah= . We can obtain a nicer constraint equation by multiplying both sides by 3, obtaining 3Ah = . Let

( , )f A h A h= + be the “objective function” we wish to minimize, and let be our constraint equation. Then

( , ) 3G A h Ah= =f Gλ∇ = ∇ at the extreme values of f on a level curve of . Calculating the gradients, and

setting up the Lagrange multiplier equations, we get: G

( ) ( )1,1 ,, & , h Af A h G A h∇ = ∇ =

11

11

A A

h h

λλ

λλ

⇒

=

=

⇒=

=

Which implies that . Substituting A h= A h= into the constraint equation, we get:

2

33

3

3

A hh hh

h

⋅ =⋅ =

=

= ±

The negative answer makes no sense in this context, so the only extreme point is 3A h= = . For this point:

( )( , ) 3, 3 3 3 2 3f A h f= = + =

This is the only extreme value for f. The function f has arbitrarily large values, for as , , in order for the volume to remain constant, equal to 1. Thus this extreme value must be the minimum we seek.

0h → A→∞

__________________________________________________________________________________

26. You need to construct a tank consisting of a cylinder with two hemispherical ends. Suppose that the material for the cylindrical portion of the tank costs $30 per square meter while the hemispherical ends cost $40 per square meter. Find the length h of the cylindrical part of the tank, and its radius , in order to enclose 10,000 m3 at minimal cost. What is the minimum cost. r Note: None of the given answers were correct for this one, so the correct answer was “n”.

The volume of this tank is 4 34103

V r 2r hπ π= = + . This is our constraint equation, and we let:

( ) 3 24, 13

G r h r r h 40π π= + =

The cost of this tank is ( ) (240 4 30 2C rπ= + )rhπ . This is our objective function, and we let:

( ) 2, 160 60f r h r rhπ π= +

Calculating the gradients, and setting up the Lagrange multiplier equations, we get:

( )

( ) 2 2

320 60 , 60 =20 16 3 , 3

4 2 , 4 2 ,

,&

,

r h r r h

r rh r r r h r

f r h

G r h

π π π π

π π π π

+ +

+ +

∇ =

∇ = =

r

( ) ( ) ( )( )

( )( )

22

20 16 3 10 16 320 16 3 2 2

2 2 260 6060

r h r hr h r r h

r r h r r hrr r

r r

ππ λπ λ

πππ λπ λ

π

+ ++ = + = =

+ +

= =⇒=

⇒

Setting the λ ’s equal, we get:

( )( ) ( )10 16 3 60 16 3 6 2 16 3 12 62

4 4 3 3

r hr h r h r h r

r r h r

r h h r

⇒+

= + = + + =+

= =

⇒

⇒ ⇒

h+

This must yield the minimum we seek, since we could find arbitrarily expensive tanks given the problem constraints. Substituting this value for into our constraint equation, we obtain (in meters): h

3 2 4 3

3

4 4 375010 10.607844179473 3

&

4 4 3750 14.1437922392943 3

r r r r

h r

π ππ

π

⇒⎛ ⎞+ = =⎜ ⎟⎝ ⎠

= =

A diagram of the tank with (close to) the optimal ratio of and h : r

The minimum cost of the tank is thus:

( )2

3 3 3

3 3 3 3 3

2

3

160 60 20 8 3

3750 3750 4 375020 8 33

3750 3750 3750 3750 375020 8 4 20 12

3750240 $84,842.88

Cost r rh r r h

Cost

π π π

ππ π π

π ππ π π π π

ππ

= + = +

⎛ ⎞= + ⋅⎜ ⎟

⎝ ⎠⎛ ⎞ ⎛

= + =⎜ ⎟ ⎜⎝ ⎠ ⎝

⎛ ⎞= ⇒ ≅⎜ ⎟

⎝ ⎠

⎞⎟⎠

__________________________________________________________________________________

27. Find the point in Octant I and on the surface 2 1xy z− = that’s closest to the origin. Let our objective function f be the function that outputs the square of the distance from an arbitrary input point ( , , )x y z to the origin:

2 2( , , ) 2f x y z x y z= + + For our constraint equation, we have 2( , , ) 1G x y z xy z= − = . Calculating the gradients, and setting up the Lagrange multiplier equations, we get:

( , , ) 2 , 2 , 2 =2 , ,&

( , , ) , , 2

f x y z x y z x y z

G x y z y x z

∇ =

∇ = −

22

22

2 2 1 or 0

xx yyyy xx

z z z

λ λ

λ λ

λ λ

= =

= =

= −⇒= −

⇒

=

⇒

Note that if x=0 then also y=0, which can’t be a point on the surface defined by our constraint equation. So we can safely assume that neither x nor y are 0. If 1λ = − , we have a contradiction, since this value substituted

into the previous two equations yields simultaneously 2y x= − and 12

y x= − , which can’t both be true

(unless x =0 which we’ve already eliminated as an option). The only remaining possibility is that z=0. Setting the first two equations for λ equal, we get:

2 22 2 x y x y x y xy x y x= = =⇒ ⇒ = ±⇒ y

Of the points ( , ,0)x x and ( , ,0)x x− , only the first is in Octant I, so we substitute this one into our constraint equation:

2 2( , , ) 1 ( , ,0) 0 1 1G x y z xy z G x x x x= ⇒ ⇒− = = − = = ±

Evaluating our objective function on the point (1 , we get ,1,0) (1,1,0) 2f = . The point we seek is (1,1,0) . __________________________________________________________________________________