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3 years ago

What is the solution to the system of equations?

1 answer:

3 0

simplify third eqn -2x+5y+2x=4
5y=4
y=4/5

substitute back to 1st n 2nd eqn n solve for x n z
-4x-4-z=18 and -2x-4-2z=12
simplify -4x-z=22 and -2x-2z=16
-4x-(-x-8)=22
-3x+8=22
x=-14/3
z=-10/3
y=4/5

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1. Find the critical points of the function. Then use the Second Derivative Test to determine whether they are local minima, loc

Nesterboy [21]

Answer:

1. the critical point are at x=0 and y=2.45, and the critical points occur at the saddle points.

2. the critical point are at x=2 and y=1, and the critical points occur at the saddle points.

Step-by-step explanation:

1.

let t= F(x,y)

Given, t= F(x,y)= 36x - 12x³ -6xy²

(1) determine $\frac{dt}{dx} \\$

and $\frac{dt}{dy} \\$ .

therefore, $\frac{dt}{dx} \\$ = 36-36x² -6y² ...................................eqn1

$\frac{dt}{dy} \\$ = -12xy .....................................eqn2

(2) for stationary points,

for eqn1, 36-36x² -6y² = 0

36x² + 6y² = 36

6x² + y² = 6

y= $\sqrt{6-6x^{2} }$ ...................................eqn3

then, substitute y in eqn2

then for, -12xy = 0

-12x $\sqrt{6-6x^{2} }$ = 0

solving this, give x=0 ...................................eqn4

substitute eqn 4 in eqn3, gives

y= $\sqrt{6}$

y= 2.45

(3) therefore, for eqn 1 and 2, the stationary points occur at x=0 and y=2.45.

(4)determine $\frac{d²t}{dx²} \\$

since, $\frac{dt}{dx} \\$ = 36-36x² -6y²

$\frac{d²t}{dx²} \\$ = 36x ...................................eqn5

(5)determine $\frac{d²t}{dy²} \\$

since, $\frac{dt}{dy} \\$ = -12xy

then $\frac{d²t}{dy²} \\$ = -12x ...................................eqn6

(6)determine $\frac{d²t}{dxdy} \\$ = $\frac{d}{dx} \\$ . $\frac{dt}{dy} \\$

= d/dx (-12xy) = -12y ...................................eqn7

substitute values of x and y (0, 2.45) in eqn 5,6,7

eqn 5 =0

eqn 6 =0

also,

eqn7 =-29.4

also for, sqr eqn7 = (-12y)² = 144y² = 864.4

(7) finally, determine whether it is maxima, minima or saddle point by:

Δ= ( $\frac{d²t}{dxdy} \\$ )² - [ $\frac{d²t}{dy²} \\$ * $\frac{d²t}{dx²} \\$ ]

for (x=0, y=2.45)

Δ = 864.4 - {(0)(0)} = 864.4

since, Δ > 0, therefore the stationary points (0,2.45) is a saddle point.

2.

let t= F(x,y)

Given, t= F(x,y)= x³ + 6xy - 6y² - 6x

(1) determine $\frac{dt}{dx} \\$

and $\frac{dt}{dy} \\$ .

therefore, $\frac{dt}{dx} \\$ = 3x² -6y-6 ...................................eqn1

$\frac{dt}{dy} \\$ = 6x - 12y .....................................eqn2

(2) for stationary points,

for eqn1, 3x² -6y-6 = 0

x² -2y -2 = 0

x² =2+2y

x =sqrt(2+2y) ...................eqn3

then, substitute x in eqn2

then for, 6x - 12y= 0

-6(sqrt(2+2y) - 12y= 0

solving this, give y= 1 and -0.5 ...................................eqn4

substitute eqn 4 in eqn3, gives

x = 2 and 1

(3) therefore, for eqn 1 and 2, the stationary points occur at x=2, y=1 and x=1, y=-0.5.

thus, the stationary points occur at (2,1) and (1, -0.5)

(4)determine $\frac{d²t}{dx²} \\$

since, $\frac{dt}{dx} \\$ = 3x² -6y-6

$\frac{d²t}{dx²} \\$ = 6x ...................................eqn5

(5)determine $\frac{d²t}{dy²} \\$

since, $\frac{dt}{dy} \\$ = 6x - 12y

then $\frac{d²t}{dy²} \\$ = -12y ...................................eqn6

(6)determine $\frac{d²t}{dxdy} \\$ = $\frac{d}{dx} \\$ . $\frac{dt}{dy} \\$

= d/dx (6x - 12y) = 6 ...................................eqn7

substitute values of x and y as (2,1) and (1, -0.5) in eqn 5,6,7

at (2,1) eqn 5 =12, eqn 6 =-12 also, eqn7 =6

at (1,-0.5) eqn 5 =6, eqn 6 =6 also, eqn7 =6

also for, sqr eqn7 = (6)² = 36

(7) finally, determine whether it is maxima, minima or saddle point by:

Δ= ( $\frac{d²t}{dxdy} \\$ )² - [ $[\frac{d²t}{dy²}$ * $\frac{d²t}{dx²} \\$ ]

for (x=2, y=1)

Δ = 36 - {(12)(-12)} = 36+144 = 180

for (x=1, y=-0.5)

Δ = 36 - {(6)(6)} = 36-36 = 0

since, Δ > 0, for (2,1) therefore the stationary points (2,1) is a saddle point.

7 0

3 years ago

Mai and priya were on scooters Mai traveled 18 meters in 4 seconds priya travels 28 meters in 6 seconds who was moving faster? P

Aleks04 [339]

Answer:

Mai

Step-by-step explanation:

We can simplify the rates they are going at.

Mai: 18/4=9/2=27/6

Priya: 28/6= 14/3= 28/6

Mai's time<Priya's time

Priya is going slower, and Mai is going faster.

6 0

3 years ago

What are the steps to simplifying radical expressions?

Assoli18 [71]

Answer:

Step 1: Find the prime factorization of the number inside the radical.

Step 2: Determine the index of the radical. In this case, the index is two because it is a square root, which means we need two of a kind.

Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the pair of 2’s and 3’s moved outside the radical.

Step 4: Simplify the expressions both inside and outside the radical by multiplying.

Step-by-step explanation:

5 0

3 years ago

What is the mean of the set of numbers? 123.4, 178.92, 0.64, 12.98, and 166.66

anzhelika [568]

Answer:

96.52

Step-by-step explanation:

You have to firstly add all the numbers. That will give you482.6. Then divide that number by the amount of numbers in the list: 482.6 / 5. That gives you 96.52.

8 0

3 years ago

The store has 28 notebooks in packs of 4. Three packs of notebooks are sold. How many packs of notebooks are left?

alukav5142 [94]

28/4=7x3=21 so 28-21=7 notebooks left

7 0

3 years ago