Algebra Equation

[JWHooper]

If x is any possible positive real number, and if the equation (8x^2) + (2/x) has the minimum value of b at x = a, then find the value of absin(pi/3). Don't use calculus -- use algebraic techniques to show solutions.

[Shalot]

If x is any possible positive real number, and if the equation (8x^2) + (2/x) has the minimum value of b at x = a, then find the value of absin(pi/3). Don't use calculus -- use algebraic techniques to show solutions.

do you know the answer? If so, you should show it because I sure don't.

[JWHooper]

Since x is any possible positive real number, we can use an algebraic technique, which is called arithmetic-geometric mean inequality.
(8x^2) + (2/x)
= 8x^2 + 1/x + 1/x
>= 3((8x^2)(1/x)(1/x))^(1/3)
= 3(8)^(1/3)
= (3)(2)
= 6
= b
Now, (8x^2) + (2/x) = 6 = b if and only if 8x^2 = 1/x (because the arithmetic-geometric inequality above shows us that 8x^2 + 1/x + 1/x >= 6)
8x^3 = 1
x^3 = 1/8
x = 1/2 = a
Therefore, absin(pi/3) = 3sin(pi/3) = (3)(sqrt(3)/2) = 3sqrt(3) / 2.

[Shalot]

Since x is any possible positive real number, we can use an algebraic technique, which is called arithmetic-geometric mean inequality.
(8x^2) + (2/x)
= 8x^2 + 1/x + 1/x
>= 3((8x^2)(1/x)(1/x))^(1/3)
= 3(8)^(1/3)
= (3)(2)
= 6
= b
Now, (8x^2) + (2/x) = 6 = b if and only if 8x^2 = 1/x (because the arithmetic-geometric inequality above shows us that 8x^2 + 1/x + 1/x >= 6)
8x^3 = 1
x^3 = 1/8
x = 1/2 = a
Therefore, absin(pi/3) = 3sin(pi/3) = (3)(sqrt(3)/2) = 3sqrt(3) / 2.


Cool. I guess. All this math hurts my brain right now. But someone definitely owes you homework help on your English assignments. :D

(welcome to LitNet btw - haven't seen you around much. math people are so cool. )

[B-Mental]

Wow, maybe you should post that on the math forums.

[papayahed]

How did you go from here:

= 8x^2 + 1/x + 1/x

to here:

>= 3((8x^2)(1/x)(1/x))^(1/3)

[kilted exile]

How did you go from here:

= 8x^2 + 1/x + 1/x

to here:

>= 3((8x^2)(1/x)(1/x))^(1/3)

Yep, I'm with papaya here,

[Taliesin]

Well, that what is called the arithmetic-geometric mean inequality which states that the arithmetic mean of positive real numbers is greater or equal than geometric mean of those numbers.
(a1+a2+a3...+an)/n >= (a1*a2*a3*....*an)^(1/n)


Using it here, where n= 3, we get:

((8x^2 + 1/x + 1/x)/3)>= ((8x^2)(1/x)(1/x))^(1/3)

[JWHooper]

But someone definitely owes you homework help on your English assignments. :D
I have American Literature in my high school currently, and I need to get the English credits in order to graduate high school. I will surely be glad if you guys would help me out.


(welcome to LitNet btw - haven't seen you around much. math people are so cool. )
Nice to meet you. I think literature people are cool too, because they are really good at reading and writing. You can add me to friends list in here, if you want.

[JWHooper]

Well, that what is called the arithmetic-geometric mean inequality which states that the arithmetic mean of positive real numbers is greater or equal than geometric mean of those numbers.
(a1+a2+a3...+an)/n >= (a1*a2*a3*....*an)^(1/n)

Great job. Also, make sure you know how to use harmonic and quadratic mean inequalities related to arithmetic and geometric mean inequality. Also, Cauchy-Schwartz inequality is important in this case.

[bluevictim]

A very clever solution to the problem! :)

But, I guess some points are lost because the problem seems rather contrived.