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never [62]
2 years ago
9

Can anyone help me solve this ASAP! /.\

Mathematics
1 answer:
maks197457 [2]2 years ago
3 0

Where is the rest of the info

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The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell
emmasim [6.3K]

Answer:

a)

Highest (-3,-3)

Lowest (2,2)

b)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

a)

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12

completing the squares:

\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of  

\bf f(x,y)=x^2+y^2

subject to the constraint

\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0

Now we have

\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)

Let \bf \lambda be the Lagrange multiplier.

The maximum and minimum must occur at points where

\bf \nabla f=\lambda\nabla g

that is,

\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y

Replacing in the constraint

\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2

from this we get

<em>x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3 </em>

<em> </em>

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

b)

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0
3 years ago
Write an equation of the line passing through the point A(8, 2) that is perpendicular to the line y = 4x−7.
Kay [80]

Answer:

y = - 1/4 + 4

Step-by-step explanation:

y = 4x - 7

Slope = 4

Slope of the perpendicular line = -1/4

Point (8,2)

(b) y-intercept: 2 - (-1/4)(8)

= 2 + 2 = 4

4 0
3 years ago
Solve the system with elimination 4x+3y=1 -3x-6y=3
Nuetrik [128]

Answer:

x = 1, y = -1

Step-by-step explanation:

If we have the two equations:

4x+3y=1 and -3x - 6y = 3, we can look at which variable will be easiest to eliminate.

y looks like it might be easy to get rid of, we just have to multiply 4x+3y=1  by 2 and y is gone (as -6y + 6y = 0).

So let's multiply the equation 4x+3y=1  by 2.

2(4x + 3y = 1)\\8x + 6y = 2

Now we can add these equations

8x + 6y = 2\\-3x-6y=3\\

------------------------

5x = 5

Dividing both sides by 5, we get x = 1.

Now we can substitute x into an equation to find y.

4(1) + 3y = 1\\4 + 3y = 1\\3y = -3\\y = -1

Hope this helped!

5 0
3 years ago
5x² + 4x - 20 = 0<br> Solve this using the quadratic formula
Natali [406]

Here's the answer, using the quadratic formula

8 0
3 years ago
Please Help. NO LINKS OR DOWNLOADS OR YOU WILL BE REPORTED.​
Aliun [14]
Did you attach a picture to this, It could be my internet but it’s not showing up on my screen
3 0
3 years ago
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