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# (pls help asap) min value of y=x^2 +16/x using mean inequality, x>0

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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!!

Apr 19, 2019

#1
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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   Apr 19, 2019
#2
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thank you so much, I really appreciate it :)

I'm just wondering, was the mean inequality used for what you just did?

And why did the x^2 turn into 2x?

hearts123  Apr 19, 2019
#3
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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 :)

Apr 19, 2019
#4
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Is the x^2 + 16 supposed to be in brakets?

Apr 20, 2019
#5
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nope, it's x^2 and 16/x added together, no brackets.

hearts123  Apr 20, 2019
#6
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ok thanks.

Are you doing a unit that might give us a hint of what you are expected to do here?

Have you done any calculus yet?

Melody  Apr 20, 2019
edited by Melody  Apr 20, 2019