2012-10-29, 02:27 AM
You can also kinda think about it this way.
If xy = 1, that makes y a function of x, ie. y = 1/x
f(x) = 5x^2 + 5/x^2
d/dx = 10x - 10/x^3
10x = 10/x^3
x^4 = 1
x = +/- 1 which is either a maximum or a minimum
d^2/dx^2 = 2 + 6/x^4
Since x^4 is always positive, the second derivative is always positive so you know there is no maximum. It follows that x = +/- 1 is always a minimum, and since y = 1/x there are minima at (1,1) and (-1,-1)
As for why it works, I see it as a way of taking one equation with two unknowns and playing with it until you have three equations with three unknowns. As it shows in my working above, you can then reduce it in to two equations with two unknowns and solve for happy fun times.
If xy = 1, that makes y a function of x, ie. y = 1/x
f(x) = 5x^2 + 5/x^2
d/dx = 10x - 10/x^3
10x = 10/x^3
x^4 = 1
x = +/- 1 which is either a maximum or a minimum
d^2/dx^2 = 2 + 6/x^4
Since x^4 is always positive, the second derivative is always positive so you know there is no maximum. It follows that x = +/- 1 is always a minimum, and since y = 1/x there are minima at (1,1) and (-1,-1)
As for why it works, I see it as a way of taking one equation with two unknowns and playing with it until you have three equations with three unknowns. As it shows in my working above, you can then reduce it in to two equations with two unknowns and solve for happy fun times.