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!!

hearts123 Apr 19, 2019

#1**+2 **

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

CPhill Apr 19, 2019

#3**+2 **

Hi Hearts123,

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]

BUT

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 :)

Melody Apr 19, 2019