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# algebra problem

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or positive real numbers x,y  and z find the minimum value of

May 19, 2019

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deleted

May 19, 2019
edited by Rom  May 19, 2019
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This is what wolframalpha has to say.

May 19, 2019
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Yeah it was easy enough to establish that 15 is the answer using calculus.

But this problem, when I looked at the other similar problems posted at about the same time,

seems to be to be solved using algebra.  As if there is a method to compare the relative

coefficients of the variables and decide based on that what the critical points would be.

I have never seen anything like and had no idea how to approach this problem w/o calculus.

Rom  May 19, 2019
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Rom can you please show me how you solve this problem using calculus?

thanks!

Guest May 19, 2019
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Critical points are found by setting the gradient equal to zero.

You then have to determine whether they are minima, maxima, or saddle points using the second derivative test.

That's not exactly trival for 3 variables but there is a method for doing it.

https://calculus.subwiki.org/wiki/Second_derivative_test_for_a_function_of_multiple_variables

Rom  May 19, 2019
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But the domain of the function is an open set (all triplets of strictly positive numbers) how do you prove that there's a minima to the function in this domain?

Guest May 19, 2019
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I am seeing a pattern that the critical points obey.

In this problem https://web2.0calc.com/questions/urgent-pls-help_1 we have

$$\left(x + 2y+4z\right)\left(\dfrac 4 x + \dfrac 2 y + \dfrac 1 z\right)= \\~\\ 4\left(x + 2y+4z\right)\left(\dfrac{1}{x}+\dfrac{1}{2y}+\dfrac{1}{4z}\right)$$

The critical point in this case occurs at

$$(x,y,z) = \left(1,\dfrac 1 2,\dfrac 1 4\right)$$

I.e. the first critical point coordinate is 1 and the ratio of the critical point coordinates is inverse of that of the ratio of the coordinates in the expression.

In the problem at hand we expect the y coordinate of the critical point will be such that

$$x^3 = 1 \Rightarrow x = 1\\~\\ 5y^3 = 1 \Rightarrow y = \dfrac{1}{5^{1/3}}\\~\\ 25z^3 = 1 \Rightarrow z = \dfrac{1}{5^{2/3}}$$

This would have been modified somehow if the denominator were not symmetric in x,y,z, but in this case it is.

Rom  May 19, 2019
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like I said above you have to apply the 3 dimensional second derivative test.

The test may prove inconclusive but under certain conditions it guarantees a minimum or maximum as the case may be.

Rom  May 19, 2019
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The second derivative test tells us whether a point is a local maximum/minimum it doesn't tell us anything about global maximum/minimum

Guest May 19, 2019