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I just want to fill the blanks 1- (B) If there is a local minimum, what is the v

ID: 3343662 • Letter: I

Question

I just want to fill the blanks


1-

(B) If there is a local minimum, what is the value of the discriminant D at that point? If there is none, type N.

(C) If there is a local maximum, what is the value of the discriminant D at that point? If there is none, type N.

(D) If there is a saddle point, what is the value of the discriminant D at that point? If there is none, type N.

First point (, ) of type
Second point (, ) of type
Third point (, ) of type
Fourth point (, ) of type

Suppose

(A) How many critical points does have in ?

(Note, is the set of all pairs of real numbers, or the -plane.)

(B) If there is a local minimum, what is the value of the discriminant D at that point? If there is none, type N.

(C) If there is a local maximum, what is the value of the discriminant D at that point? If there is none, type N.

(D) If there is a saddle point, what is the value of the discriminant D at that point? If there is none, type N.

(E) What is the maximum value of on ? If there is none, type N.

(F) What is the minimum value of on ? If there is none, type N.

Explanation / Answer

1.

By discriminant, they are referring to the discriminant of the Hessian, or second derivatives

fx = 2x-10

fy = 2y - 2

fxx = 2

fyy = 2

fxy = fyx = 0

The determinant is

2 0

0 2

Thus, the discriminant is 4.

The eigenvalues are both 2, so this is a positive definite matrix and we have a minimum (not asked for)

The discriminant is 4 everywhere, including at the minimum

The derivative equals 0 at 2x - 10 = 0 and 2y - 2 = 0, or (5, 1), so this is the minimum.

x^2 + y^2 - 10x - 2y + 4 =

5^2 + 1^2 - 10*5 - 2*1 + 4 =

25 + 1 - 50 - 2 + 4 = -22


2. f(x,y) = xy(1-7x-2y)=xy -7x^2y -2xy^2

fx = y - 14xy -2y^2 = y(1 - 14x - 2y)

fy = x - 7x^2 -4xy = x(1 - 7x - 4y)

fxx = -14y

fyy = -4x

fxy = 1 - 14x -4y


The Hessian is

-14y 1 - 14x -4y

1 - 14x -4y -4x


fx=y(1 - 14x - 2y)

fy=x(1 - 7x - 4y)


We have (0, 0) as a solution.

As the Hessian is

0 1

1 0

the discriminant is -1, so we have a saddle point.


If x = 0 and y is not equal to 0, then (1-14(0) -2y) = 0

1 = 2y

y = 1/2


The discriminant is

-14(1/2) 1 - 14(0) -4(1/2)

1 - 14(0) -4(1/2) 0

or

-7 -1

-1 0

The discriminant is -1, so this is a saddle point.



If y = 0 and x is not equal to 0, then

1 - 7x -4(0) = 0

1 = 7x

x = 1/7


The Hessian is

-14(0) 1 - 14(1/7) -4(0)

1 - 14(1/7) -4(0) -4(1/7)

or

0 -1

-1 -4/7

The discriminant is -1, so this is a saddle point.


If neither equals 0

1 - 14x - 2y = 0

1 - 7x - 4y = 0

14x + 2y = 1

7x + 4y = 1

Double the bottom equation

14x + 8y = 2

Subtract the top equation from the bottom equation.

6y = 1

y = 1/6

7x + 4y = 1

7x = 1 - 4(1/6)

7x = 1/3

x = 1/21

The Hessian is

-14y 1 - 14x -4y

1 - 14x -4y -4x

-14(1/6) 1 - 14(1/21) - 4(1/6)

1 - 14(1/21) - 4(1/6) -4(1/21)

or

-7/3 -1/3

-1/3 -4/21


The discriminant is (-7/3)(-4/21)-(-1/3)^2 = 28/63 - 1/9 = 4/9 - 1/9 = 1/3

The two diagonal elements are negative.

Thus, our matrix is negative definite and this is a maximum.

0

0

S

0

1/2

S

1/21

1/6

MA

1/7

0

S


3.x^7y - 7x^6 + y

fx = 7x^6y - 42x^5

fy = x^7 + 1


x^7 + 1 = 0, so x^7 = -1, or x = -1

Then, 7x^6y - 42x^5 = 0, so

7y + 42 = 0

7y = -42

y = -6


fxx = 42x^5y - 210x^4

fyy = 0

fxy = 7x^6

The Hessian is

42x^5y - 210x^4 7x^6

7x^6 0

or

42 7

7 0

The discriminant is -49, so this is a saddle point

-1

-6

S


Everything else is left blank


A linear function contained within a polygon has its maxima and minima at the vertices.

20 - 4x + 9y at

(0, 0), (9, 0), (9, 13)

f(0,0) = 20 - 4(0)+ 9(0) = 20

f(9,0) = 20 - 4(9)+ 9(0) = -16

f(9, 13) = 20 - 4(9)+ 9(13) = 101


Min is -16 at (9,0) and max is 101 at (9, 13)

-16

9

0

101

9

13


5.

Use LaGrange multipliers

l(x,y) = 6x + y + L(x^2+9y^2 - 1)

Gradient is

(6+2Lx, 1 + 18Ly)

2Lx = -6

18Ly = -1

Dividing,

x/9y = 6

x = 54y


Substitute into

x^2 + 9y^2 = 1

(54y)^2 + 9y^2 = 1

(18*3y)^2 + (3y)^2 = 1

325*9y^2 = 1

y^2 = 1/(9*325)

y = plus or minus 1/15 sqrt(13)

x = 54 y = 54 * plus or minus 1/15 sqrt(13) = plus or minus 18/5 sqrt(13)


Then, 6x + y = plus or minus 6 * 18/5 sqrt(13) + 1/15 sqrt(13) =

plus or minus 65/3 sqrt(13)


Maximum value

65/(3 sqrt(13)) = 6.00925212577332


Minimum value

-65/(3 sqrt(13)) = -6.00925212577332