what's the minimum value of y=x^2 +16/x ? x > 0
I have to use the mean inequality to get the answer: (a+ b)/2 > sqrt(a*b)
(it's actually greater or equal to, but if I substitute (x, y) as the (a, b) values, it's just > because x and y aren't equal)
so I have (x + y)/ 2 > sqrt(x*y)
I am not supposed to solve this question graphically, or with a graphing calculator.
Any help is appreciated!!
y = x^2 + 16x^(-1)
Take the derivative....set to 0 and solve
y ' = 2x - 16/x^2 = 0
2x^3 - 16 = 0
x^3 - 8 = 0
Taking the positive cube root produces x = 2
And the minimum value is ( 2)^2 + 16/2 = 4 + 8 = 12
Chris has answered it using calculus. Have you done any calculus yet?
That is the normal method I would use too.
[The differential of x^2 = 2x]
I do not think that is what you want.
The trouble is that neither of us have worked out the relevance of that inequality (which is always true by the way).
I will think about it some more :)
Is the x^2 + 16 supposed to be in brakets?