Quadratic function minimal value

[YoungPhysicist]

Homework Statement

##x,y,z \in \mathbb{R}##, find the minimal value for $$x^2+2y^2+z^2-6x+4y-10z+17$$

Homework Equations

None

The Attempt at a Solution

First I try to use”complete the square” method to make the polynomial something like:
$$(x-3)^2+(\sqrt2y+\sqrt2)^2+(z-5)^2-19$$
Then I am stuck.

Note:This problem is on a univariate quadratic worksheet,so it should be able to be solved using nothing more than that.

 

[fresh_42]

If you inspect the formula, what do you see?

 

[YoungPhysicist]

If you inspect the formula, what do you see?

I think I get it.Is the answer -19?

 

[fresh_42]

I think I get it.Is the answer -19?

That's the value at the minimum, but the question is about its coordinates. Where is it ##f(x,y,z)=-19## and why is it a minimum? I meant this question "What do you see" quite literally.

 

[YoungPhysicist]

but the question is about its coordinates.

I don’t quite get this part.I already answer the question.Why should I focus on the coordinates?

when x = 3, y = -1, z = 5, the entire formula will be eliminated except -19, which is the minimal value of the function.Is that what you meant?

 

[fresh_42]

I don’t quite get this part.I already answer the question.Why should I focus on the coordinates?

when x = 3, y = -1, z = 5, the entire formula will be eliminated except -19, which is the minimal value of the function.Is that what you meant?

Almost. Where is the minimum? means to name ##x = 3, y = -1, z = 5##, not ##-19##, which would have been: What is the value at the minimum? which wasn't asked. So it is the essential part of an answer, not the version you gave first.

Why is it the minimum? Because all other values contribute positive amounts, as all squares are non-negative. This is the reason, which also should be part of the answer.

What do you see? isn't really necessary to answer, but a good exercise. What we have here is an ellipsoid, a squeezed ball! It also is placed along the coordinate axis, so the minimum will be where it "touches ground". That means we have a global minimum and a unique one.

All these comments I made should be part of a good answer. ##-19## is neither an answer nor good.

 

[YoungPhysicist]

So a good answer will be:

Since all squares can only give non negative numbers,when x = 3,y = -1 ,z = 5, all the squares will give “0”, which is the smallest value it can offer, so its constant term -19 will ben the minimum.

 

[YoungPhysicist]

What do you see? isn't really necessary to answer, but a good exercise. What we have here is an ellipsoid, a squeezed ball! It also is placed along the coordinate axis, so the minimum will be where it "touches ground". That means we have a global minimum and a unique one.

So do you mean the function shapes like a ellipse? Then how can it be a function?

 

[fresh_42]

I don't see a function, only an algebraic expression. But if we wrote ##f(x,y,z)=(x-3)^2+(\sqrt 2 y + \sqrt 2 )^2+(z-5)^2-19## we would get only one image point at each location ##(x,y,z)##, so it is a function. Its graph is four dimensional. It is only then no function, if we want to write it as, e.g. ##z= \ldots## However, to do so, we need an equation, which I do not see we have.

As I said it is an ellipsoid, I meant it's level sets: ##f(x,y,z)-19 = c## for a certain level ##c##. These points solve an equation and the result looks like a football, e.g. for ##c=0##: https://www.wolframalpha.com/input/?i=(x-3)^2+(root(2)y+root(2))^2+(z-5)^2=19

 

[Ray Vickson]

That's the value at the minimum, but the question is about its coordinates. Where is it ##f(x,y,z)=-19## and why is it a minimum? I meant this question "What do you see" quite literally.

Actually, the OP is correct; the question asks for the "minimal value for ... ", which is the -19 in this case. Nevertheless, you are correct in pointing out to the OP that a more meaningful answer would be something like "the minimum value of ##f## is -19, and occurs at ##(3,-1,5).## Perhaps the person setting the question should have done a better job.

 

[fresh_42]

Actually, the OP is correct; the question asks for the "minimal value for ...

Oops, and sorry @Young physicist. Quite a rare question.

 

[YoungPhysicist]

Oops, and sorry @Young physicist. Quite a rare question.

Nah.nothing.I learn a lot more things in the process than just the minimal value.Thank you @fresh_42