SOLUTIONS - UCSD Mathematics

SOLUTIONS

Problem 1. Find the critical points of the function

f (x, y) = 2x3 - 3x2y - 12x2 - 3y2

and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Solution: Partial derivatives fx = 6x2 - 6xy - 24x, fy = -3x2 - 6y.

To find the critical points, we solve fx = 0 = x2 - xy - 4x = 0 = x(x - y - 4) = 0 = x = 0 or x - y - 4 = 0 fy = 0 = x2 + 2y = 0.

When x = 0 we find y = 0 from the second equation. In the second case, we solve the system below by substitution

x - y - 4 = 0, x2 + 2y = 0 = x2 + 2x - 8 = 0

= x = 2 or x = -4 = y = -2 or y = -8.

The three critical points are

(0, 0), (2, -2), (-4, -8).

To find the nature of the critical points, we apply the second derivative test. We have

A = fxx = 12x - 6y - 24, B = fxy = -6x, C = fyy = -6. At the point (0, 0) we have

fxx = -24, fxy = 0, fyy = -6 = AC - B2 = (-24)(-6) - 0 > 0 = (0, 0)is local max .

Similarly, we find since and since

(2, -2) is a saddle point AC - B2 = (12)(-6) - (-12)2 =< 0

(-4, -8) is saddle AC - B2 = (-24)(-6) - (24)2 < 0.

The function has no global min since lim f (x, y) = -

y,x=0

and similarly there is no global maximum since lim f (x, y) = .

x,y=0

1

Problem 2.

Determine the global max and min of the function

f (x, y) = x2 - 2x + 2y2 - 2y + 2xy

over the compact region

-1 x 1, 0 y 2.

Solution: We look for the critical points in the interior: f = (2x - 2 + 2y, 4y - 2 + 2x) = (0, 0) = 2x - 2 + 2y = 4y - 2 + 2x = 0 = y = 0, x = 1.

However, the point (1, 0) is not in the interior so we discard it for now. We check the boundary. There are four lines to be considered:

? the line x = -1:

f (-1, y) = 3 + 2y2 - 4y.

The critical points of this function of y are found by setting the derivative to zero:

(3 + 2y2 - 4y) = 0 y

=

4y - 4 = 0

=

y = 1 with

f (-1, 1) = 1 .

? the line x = 1:

f (1, y) = 2y2 - 1.

Computing the derivative and setting it to 0 we find the critical point y = 0. The corresponding point (1, 0) is one of the corners, and we will consider it separately below.

? the line y = 0:

f (x, 0) = x2 - 2x.

Computing the derivative and setting it to 0 we find 2x - 2 = 0 = x = 1. This gives the

corner (1, 0) as before.

? the line y = 2:

f (x, 2) = x2 + 2x + 4

with critical point x = -1 which is again a corner.

Finally, we check the four corners (-1, 0), (1, 0), (-1, 2), (1, 2).

The values of the function f are

f (-1, 0) = 3 , f (1, 0) = -1 , f (-1, 2) = 3 , f (1, 2) = 7 .

From the boxed values we select the lowest and the highest to find the global min and global max. We conclude that

global minimum occurs at (1, 0)

global maximum occurs at (1, 2).

Problem 3. Using Lagrange multipliers, optimize the function

f (x, y) = x2 + (y + 1)2

subject to the constraint

2x2 + (y - 1)2 18.

Solution: We check for the critical points in the interior

fx = 2x, fy = 2(y + 1) = (0, -1) is a critical point .

The second derivative test shows this a local minimum with

fxx = 2, fyy = 2, fxy = 0 f (0, -1) = 0 .

We check the boundary

g(x, y) = 2x2 + (y - 1)2 = 18

via Lagrange multipliers. We compute

f = (2x, 2(y + 1)) = g = (4x, 2(y - 1)).

Therefore

1 2x = 4x = x = 0 or = 2

2(y + 1) = 2(y - 1).

In the first case x = 0 we get

g(0, y) = (y - 1)2 = 18 = y = 1 + 3 2, 1 - 3 2

with values

f (0, 1 + 3 2) = (2 + 3 2)2 , f (0, 1 - 3 2) = (2 - 3 2)2 .

In

the

second

case

=

1 2

we

obtain

from

the

second

equation

2(y + 1) = y - 1 = y = -3.

Now,

g(x, y) = 18 = x = ?1.

At (?1, -3), the function takes the value

f (?1, -3) = (?1)2 + (-3 + 1)2 = 5.

By comparing all boxed values, it is clear the (0, -1) is the minimum, while (0, 1 + 3 2) is the maximum.

Problem 4.

Consider the function

w = ex2y

where

1

x = u v, y = uv2 .

Using the chain rule, compute the derivatives

w w ,.

u v

Solution: We have

w x

=

2xy exp(x2y)

=

1 2u v uv2

exp

u2v

?

1 uv2

2

u

= v3/2 exp v

w = x2 exp(x2y) = u2v exp u

y

v

x x u

= v, =

u

v 2 v

Thus Similarly,

y

1 y

2

u = - u2v2 , v = - uv3 .

w w x w y 2 = ? + ? = exp

u x u y u v3/2

u v

?

v

-

u2v

exp

u v

1 ? u2v2 =

2

u1

u1

u

= exp - exp = exp .

v

vv

vv

v

w w x w y u

u

u = x ? u + y ? u = - v2 exp v .

Problem 5.

(i) For what value of the parameter a, will the planes ax + 3y - 4z = 2, x - ay + 2z = 5

be perpendicular? (ii) Find a vector parallel to the line of intersection of the planes

x - y + 2z = 2, 3x - y + 2z = 1.

(iii) Find the plane through the origin parallel to z = 4x - 3y + 8.

(iv) Find the angle between the vectors v = (1, -1, 2), w = (1, 3, 0).

(v) A plane has equation For what values of a is the vector normal to the plane?

z = 5x - 2y + 7. 1

(a, 1, ) 2

Solution:

(i) The normal vectors to the two planes are n1 = (a, 3, -4), n2 = (1, -a, 2).

The planes are perpendicular if n1, n2 are perpendicular. We compute the dot product n1 ? n2 = 0 = a ? 1 + 3 ? (-a) + (-4) ? 2 = 0 = -2a - 8 = 0 = a = -4.

(ii) The vectors normal to the two planes are n1 = (1, -1, 2), n2 = (3, -1, 2).

The line of intersection will be perpendicular to both n1, n2. But so is the cross product. Thus the line of intersection will be parallel to the cross product

n1 ? n2 = (1, -1, 2) ? (3, -1, 2) = (0, 4, 2). (iii) The second plane must have the same normal vector hence the same coefficients for x, y, z.

Since it passes through the origin, the equation is z = 4x - 3y.

(iv) We compute the angle using the dot product

v?w

-2

1

cos =

= = - .

||v|| ? ||w|| 6 ? 10

15

(v) The plane has the equation

5

17

5x - 2y - z = -7 = - x + y + z =

2

22

hence

a

normal

vector

is

(-

5 2

,

1,

1 2

).

Comparing

with

the

vector

we

are

given,

we

see

that

5 a=- .

2

Problem 6.

(i) Compute the second degree Taylor polynomial of the function

f (x, y) = ex2-y

around (1, 1). (ii) Compute the second degree Taylor polynomial of the function

around x = .

f (x) = sin(x2)

(iii) The second degree Taylor polynomial of a certain function f (x, y) around (0, 1) equals

1 - 4x2 - 2(y - 1)2 + 3x(y - 1).

Can the point (0, 1) be a local minimum for f ? How about a local maximum?

Solution:

(i) After computing all derivatives and substituting, we find the answer

1

+

2(x

-

1)

-

(y

-

1)

+

3(x

-

1)2

+

1 (y

-

1)2

-

2(x

-

1)(y

-

1).

2

(ii) We have f ( ) = 0. The first derivative is

fx = 2x cos x2

=

fx( ) = 2 cos = -2 .

The second derivative is

fxx = 2 cos x2 - 2x sin x2

=

fxx( ) = -2.

The Taylor polynomial is

-2 (x

-

)

-

(x

-

)2

=

-x2

+

.

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