Find points on a plane at(0,0,0) 2x+3y+yz=12

[Unemployed]

Homework Statement

Find points on a plane at(0,0,0) 2x+3y+yz=12

Homework Equations

d^2=x^2 +y^2-(2x-3y-12)/y

The Attempt at a Solution

fx=2x-(4 (2x-3y-12)/y)
fxx= 2-8/y
Fy 2y-(2*3*(2x-3y-12)/y)
Fyy= (4y+3)/y^2

I am getting round numbers to plug back in. Totally wrong approach?

 

[Unemployed]

Find points on a plane (2x+3y+yz=12) at 0,0,0

Homework Statement

Find points on a plane 2x+3y+yz=12 at (0,0,0)

Homework Equations

d^2=x^2+y^2+((2x+3y-12)/-y)^2

The Attempt at a Solution

My calculation: fx= 2xy-8x+12y-48/y
Looked on a derivative calculator: (2xy^2+8X+12y-48/)y^2

My calc:
fy= (^33-2X-3Y+12)/y^2

The fyy and fxx get convoluted assuming i did the first derivatives correctly. Then the d-test for saddle. Need a comprehensive explanation of this example. Thanks

 

[Mark44]

Homework Statement

Find points on a plane 2x+3y+yz=12 at (0,0,0)

This is not the equation of a plane. Are you trying to find points on the plane that is tangent to the surface 2x+3y+yz=12 at (0,0,0)?

The problem with that interpretation of what you wrote is that (0, 0, 0) is not a point on this surface.

Homework Equations

d^2=x^2+y^2+((2x+3y-12)/-y)^2

The Attempt at a Solution

My calculation: fx= 2xy-8x+12y-48/y
Looked on a derivative calculator: (2xy^2+8X+12y-48/)y^2

My calc:
fy= (^33-2X-3Y+12)/y^2

The fyy and fxx get convoluted assuming i did the first derivatives correctly. Then the d-test for saddle. Need a comprehensive explanation of this example. Thanks

How about starting with the exact wording of this problem.

 

[Unemployed]

How is it not a plane? I graphed on a 3-D graphing calculator (using what i got as z) and it was a plane, it was twisted but a plane. Planes usually take the formula a(x1-xo)+b(y1-yo)+c(z1-z0)=Constant

The onlyway that this equation is incorrect is it was supposed to be 2x+3y+4z=12

Would that make more sense?

 

[Mark44]

How is it not a plane? I graphed on a 3-D graphing calculator (using what i got as z) and it was a plane, it was twisted but a plane.

There is no twist in a plane.

Planes usually take the formula a(x1-xo)+b(y1-yo)+c(z1-z0)=Constant

Sort of. It should be a(x - x0) + b(y - y0) + c(z - z0) = Constant.

The onlyway that this equation is incorrect is it was supposed to be 2x+3y+4z=12

Would that make more sense?

Well, at least that's actually a plane. The problem is that it doesn't go through (0, 0, 0), which is pretty easy to check.

What is the actual wording of the problem you're trying to solve?

 

[Unemployed]

These are problems given in notes as HW made up as the professor goes along rather than straight out of the book. You are right in that I am not sure in what the hell he is asking.

I assume the plane is 2x+3y+4z=12
Then it says:
Find the point on the plane and the point (0,0,0)?

Following the example:
Classify the critical points and max, min,saddle or test fails

x+y+z=3

at (1,2,3)

 

[HallsofIvy]

Does it say "Find the point on the plane 2x+ 3y+ 4z= 12 closest to (0, 0, 0)"?

It that were the problem, you would want to find the line through (0, 0, 0) perpendicular to 2x+ 3y+ 4z= 12 and find where that line crosses the plane.

 

[Mark44]

Following the example:
Classify the critical points and max, min,saddle or test fails

x+y+z=3

at (1,2,3)

(1, 2, 3) is not a point on the plane x + y + z = 3.

 

[Unemployed]

it still does say find point on the plane 2x+3y+4z+12 and the point (0,0,0)

But I am guessing he meant find the point closest to 0,0,0

The prior example x+y+z=3

at point (1,2,3) does show use of the perpendicular. Thanks for the help deciphering

 

[HallsofIvy]

To find the point on x+ y+ z= 3, closest to (1, 2, 3), note that the vector <1, 1, 1> is normal to the plane and so x= 1+ t, y= 2+ t, z= 3+ t is perpendicular to the plane. That point intersects the plane when (1+ t)+ (2+ t)+ (3+ t)= 6+ 3t= 3 so 3t= -3 and t= -1.

That is the point x= 1- 1= 0, y= 2- 1= 1, z= 3- 1= 2 or (0, 1, 2).

 

[Mark44]

So minimizing the square of the distance (which automatically minimizes the distance), you have d² = f(x, y) = x² + y² + (3 - x/2 - 3y/4)².

Now you can find your two partial derivatives and set them to zero to find the x, y, and z coordinates of the point on the plane that is closest to the origin.

 

[HallsofIvy]

So minimizing the square of the distance (which automatically minimizes the distance), you have d² = f(x, y) = x² + y² + (3 - x/2 - 3y/4)².

Now you can find your two partial derivatives and set them to zero to find the x, y, and z coordinates of the point on the plane that is closest to the origin.

This works but is not the way I would do it. For planes (and linear problems in general) it is not necessary use Calculus. Of course, if the problem were your original 2x+ 3y+ yz= 12, which is not a plane, you would have to use Calculus but apparently that was a mistype for 2x+ 3y+ 4z= 12.

 

[Mark44]

I wouldn't do it this way, either, but I was following the path the OP started along.