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Range of 3d function? 
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#1
Sep1204, 11:45 AM

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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
Sep1204, 11:54 AM

P: 1,367

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
Sep1204, 11:59 AM

P: n/a

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
Sep1204, 12:06 PM

P: n/a

Range of 3d function?
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
Sep1204, 12:37 PM

P: 1,367

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



#6
Sep1204, 12:44 PM

P: n/a

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



#7
Sep1204, 02:08 PM

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[tex]x(u,v)=u\cos\thetav\sin\theta[/tex] [tex]y(u,v)=u\sin\theta+v\cos\theta[/tex] Hence we have: [tex]x^{2}+xy+y^{2}=u^{2}+v^{2}+\frac{u^{2}v^{2}}{2}\sin(2\theta)+uv\cos(2\theta)[/tex] By choosing [tex]\theta=\frac{\pi}{4}[/tex], we may rewrite f in terms of u and v like this: [tex]f=\frac{3}{2}u^{2}+\frac{1}{2}v^{2}+\fsqrt{2}(uv)2\sqrt{2}(u+v)[/tex] Furthermore: [tex]f=\frac{3}{2}(u^{2}\frac{2\sqrt{2}}{3}u)+\frac{1}{2}(v^{2}6\sqrt{2}v)[/tex] Or: [tex]f=\frac{3}{2}(u\frac{\sqrt{2}}{3})^{2}\frac{2}{3}+\frac{1}{2}(v3\sqrt{2})^{2}9[/tex] Finally: [tex]f=\frac{3}{2}(u\frac{\sqrt{2}}{3})^{2}+\frac{1}{2}(v3\sqrt{2})^{2}\frac{29}{3}[/tex] Clearly, we must have: [tex]f\geq\frac{29}{3}[/tex] 


#8
Sep1204, 03:53 PM

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


#9
Sep1304, 02:10 AM

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


#10
Sep1304, 02:14 AM

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#11
Sep1304, 02:19 AM

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OK, I was looking at the ffunction 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 axisrotation technique proposed: Given: [tex]Ax^{2}+2Bxy+Cy^{2}[/tex] We have, introducing u and v: [tex]Du^{2}+uv(2B\cos(2\theta)+(CA)\sin(2\theta))+Ev^{2}[/tex] Or, in general, we may use: [tex]tan(2\theta)=\frac{2B}{AC}[/tex] to eliminate the crossproduct term. The signs of D and E will then determine the range. 


#12
Sep1304, 04:58 AM

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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. [tex] \mbox{ If } A\geq 0 \mbox{ and } C \geq 0 \mbox{ than } f \geq 0[/tex] [tex] \mbox{ If } A\leq 0 \mbox{ and } C \leq 0 \mbox{ than } f \leq 0[/tex] 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 [tex]f(x,y)=A[x^2+\frac{2B}{A}xy+\frac{C}{A}y^2]=A[(x+\frac{B}{A}y)^2+\frac{CAB^2}{A^2}y^2]=[/tex] [tex]=A(x+\frac{B}{A}y)^2+(C\frac{B^2}{A})y^2[/tex] Now you can apply the previous argument. If [tex] A> 0 \mbox{ and } C\frac{B^2}{A} \geq 0 \mbox{ then } f(x,y) \geq 0[/tex] if [tex] A< 0 \mbox{ and } C\frac{B^2}{A} \leq 0 \mbox{ then } f(x,y) \leq 0 [/tex] 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 


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