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Equation with x,y as exponential functions 
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#19
Mar1712, 04:00 AM

P: 44




#20
Mar1712, 08:43 AM

HW Helper
P: 3,531

I've just been playing with things for a bit, and just thought I'd dumb things down for simplicity, which could also mean I dumbed things down a little too far.
If we look at the exponents only, and take the addition of them we have a function [tex]f(x,y)=x^2+y+y^2+x[/tex] [tex]=\left( x+\frac{1}{2} \right)^2+\left( y+\frac{1}{2} \right)^2\frac{1}{2}[/tex] So we clearly have a symmetry about y=x (as we already knew) and the minimum value f can take is 1/2. At this point [itex](x,y)=(1/2,1/2)[/itex] which is where the minimum of f occurs, we found that [tex]g(x,y)=16^{x^2+y}+16^{x+y^2}=1[/tex] And so if we take some other value, say, [itex]x=\frac{1}{2}+k[/itex], [itex]y=\frac{1}{2}k[/itex], [itex]k\neq0[/itex] *** then with these values we have [tex]x^2+y=\left(k\frac{1}{2}\right)^2\frac{1}{2}k[/tex] [tex]=\left(k1\right)^2\frac{5}{4}[/tex] [tex]y^2+x=\left(k+\frac{1}{2}\right)^2\frac{1}{2}+k[/tex] [tex]=\left(k+1\right)^2\frac{5}{4}[/tex] Therefore, g becomes [tex]g(k)=16^{(k1)^25/4}+16^{(k+1)^25/4}[/tex] [tex]=\frac{1}{32}\left(16^{(k1)^2}+16^{(k+1)^2}\right)[/tex] And for all k [tex]16^{(k1)^2}+16^{(k+1)^2}>16+16=32[/tex] So we only have the minimum at (1/2,1/2) which means this is the only combination (x,y) that satisfies [itex]g(x,y)=1[/itex] *** I'm uneasy about making this assumption that if we take some small linear increase of k in one variable, we should decrease the other variable with the same magnitude k. 


#21
Mar1712, 11:33 AM

P: 44

Holy Menials! Mentallic, you're a genius!!!
I love you! I just thought of something new (credits go to you of course) We know that, Arithmetic mean >= geometric mean. Thus, [16^(x*x + y) + 16^(y*y + x)]/2 >= { [16^(x*x + y)]*[16^(y*y + x)] } ^1/2 Hence g(x,y) >= 2*[16^( x*x + y*y + x + y)]^1/2 Now (x*x + y*y + x + y) >= 1/2 Thus g(x,y) >= 2 *[16^(1/4)] And g(x,y)>= 1 But at (x,y) = (1/2,1/2) g(x,y)=1 SOLVED!!!! I LOVE YOU!!!! 


#22
Mar1712, 06:37 PM

HW Helper
P: 3,531

Oh awesome! You found the answer without using any crude assumptions like I did
And I'm hardly a genius, I just gave you the kick start you needed. Using the geometric mean <= arithmetic mean was the real genius here 


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