# Range of 3d function?

1. ### Khan

0
I'm not sure how to approach this problem:

Find the range of f(x,y) = Ax^2 + 2Bxy + Cy^2 (Answer will be in terms of A, B, and C)

Thanks! And we haven't yet been allowed to use partial derivatives, so they can't be used to solve it.

2. ### e(ho0n3

This question is rather vague. What is the domain of f? And why do you say the answer will be in terms of A, B, and C? Maybe you're asking for something else?

3. ### Khan

0
I agree that it's vague, but that's how the question was given to me. I assume the domain is all real numbers, and wouldn't the range have to depend on A, B, and C?

4. ### Khan

0
It may be helpful if you helped me with a different question, and maybe then I could figure this one out on my own:

find the range of f(x,y) = x^2 + xy + y^2 + 2x - 4y over the domain of all reals

5. ### e(ho0n3

OK. I understand now. Basically you're asked to find the global extremum. Are you really not allowed to use partial derivatives?

6. ### Khan

0
Yeah we really aren't. We're supposed to use algebra or single variable calculus only.

7. ### arildno

12,015
1.Eliminate the cross-product term in the following manner:
$$x(u,v)=u\cos\theta-v\sin\theta$$

$$y(u,v)=u\sin\theta+v\cos\theta$$
Hence we have:
$$x^{2}+xy+y^{2}=u^{2}+v^{2}+\frac{u^{2}-v^{2}}{2}\sin(2\theta)+uv\cos(2\theta)$$
By choosing $$\theta=\frac{\pi}{4}$$, we may rewrite f in terms of u and v like this:
$$f=\frac{3}{2}u^{2}+\frac{1}{2}v^{2}+\fsqrt{2}(u-v)-2\sqrt{2}(u+v)$$
Furthermore:
$$f=\frac{3}{2}(u^{2}-\frac{2\sqrt{2}}{3}u)+\frac{1}{2}(v^{2}-6\sqrt{2}v)$$
Or:
$$f=\frac{3}{2}(u-\frac{\sqrt{2}}{3})^{2}-\frac{2}{3}+\frac{1}{2}(v-3\sqrt{2})^{2}-9$$
Finally:
$$f=\frac{3}{2}(u-\frac{\sqrt{2}}{3})^{2}+\frac{1}{2}(v-3\sqrt{2})^{2}-\frac{29}{3}$$
Clearly, we must have:
$$f\geq-\frac{29}{3}$$

8. ### Tide

3,143
Why not just transform to polar coordinates?
$$f(x, y) = f(r, \theta) = r^2 \left(A \cos^2 \theta + B \cos \theta \sin \theta + C \sin^2 \theta \right)$$
It should be obvious then what the range is.

9. ### arildno

12,015
You've forgotten the linear terms, Tide!
It might well be that polar coordinates provide simpler expressions..

10. ### Tide

3,143
First, there aren't any linear terms so you're probably talking about the cross terms! I didn't forget them! What I did was to accidentally drop the factor of two in the coefficient of xy. Thanks for pointing me to my error!

11. ### arildno

12,015
OK, I was looking at the f-function given in post 4, not the one given in post 1!
While polar coordinates might be instructive in post 1, it's also quite simple to use the axis-rotation technique proposed:
Given:
$$Ax^{2}+2Bxy+Cy^{2}$$
We have, introducing u and v:
$$Du^{2}+uv(2B\cos(2\theta)+(C-A)\sin(2\theta))+Ev^{2}$$
Or, in general, we may use:
$$tan(2\theta)=\frac{2B}{A-C}$$
to eliminate the cross-product term.
The signs of D and E will then determine the range.

Last edited: Sep 13, 2004
12. ### ehild

11,782
First investigate the cases when one or more coefficients are zero. This is relatively easy. If all are zero the range is zero. If B is not zero but both A and C are zero, f can take any real values.

Now we assume that B=0 and at least one of A and B is not zero.

If B=0, f(x,y) = Ax^2+Cy^2.

$$\mbox{ If } A\geq 0 \mbox{ and } C \geq 0 \mbox{ than } f \geq 0$$

$$\mbox{ If } A\leq 0 \mbox{ and } C \leq 0 \mbox{ than } f \leq 0$$

If A<0, B>0 or A>0, B<0 f can be either positive or negative, and can have any values in the range of real numbers.

Now assume that A is not zero. Then you can rewrite the function as

$$f(x,y)=A[x^2+\frac{2B}{A}xy+\frac{C}{A}y^2]=A[(x+\frac{B}{A}y)^2+\frac{CA-B^2}{A^2}y^2]=$$

$$=A(x+\frac{B}{A}y)^2+(C-\frac{B^2}{A})y^2$$

Now you can apply the previous argument.

If

$$A> 0 \mbox{ and } C-\frac{B^2}{A} \geq 0 \mbox{ then } f(x,y) \geq 0$$

if

$$A< 0 \mbox{ and } C-\frac{B^2}{A} \leq 0 \mbox{ then } f(x,y) \leq 0$$

if the sign of both coefficients are different f can take any real values.

If A=0 but C is not, you can arrive to the range by factoring out C first.

ehild