Now first looking at the terms, the first thing I instinctively did was, I took y=x+1..so now the equation reduces to 1+16/(x*(x+1))+6(2x+1)..The function will be increasing in the interval (0,1) and after 1. So the minimum is obtained at x=1 which is 27
minimum value occurs..when x=2 y=1 or y=2 x=1... the expression equals 27...for any other positive value the expression is greater than this.... Let x=ky 1/(y^2(k-1)^2)+ 16/(K*y^2)+ 6y(k+1) we can try substituing values for K and Y....
I dont think this is best approach...just trial and error
be distinct positive reals. Determine the minimum value of 
and determine when that minimum occurs.
Now first looking at the terms, the first thing I instinctively did was, I took y=x+1..so now the equation reduces to 1+16/(x*(x+1))+6(2x+1)..The function will be increasing in the interval (0,1) and after 1. So the minimum is obtained at x=1 which is 27
ReplyDeleteOops interchanged x and y..I considered 1/(y-x)^2
ReplyDeleteminimum value occurs..when x=2 y=1 or y=2 x=1...
ReplyDeletethe expression equals 27...for any other positive value the expression is greater than this....
Let x=ky
1/(y^2(k-1)^2)+ 16/(K*y^2)+ 6y(k+1)
we can try substituing values for K and Y....
I dont think this is best approach...just trial and error