Why isn't it y^4?

Algebra Level 5

$\displaystyle\sqrt{xy} +\sqrt{yz} +\sqrt{xz} +x^4+y^3+z^4$

What is the minimum value of the above expression if $xyz=2$ ? Give your answer to 1 decimal place.

Clarification : Yes, it's $y^3$ and not $y^4$ .

The answer is 10.6.

2 solutions

Feathery Studio
May 12, 2015

I'm not sure if this is right, but the traditional minimum for each value is when all three of them are equal, or in this case, $2^{1/3}$ . However, since $y$ has a lesser exponent than the other variables, we can maximize it and minimize the other two. So we can make $x = 2^{1/6}$ , $y=2^{2/3}$ , and $z=2^{1/6}$ . It meets the condition that $xyz = 2$ , and it gives us a value of $\boxed{10.6}$ .

Moderator note:

This solution does not provide a proof of the minimum. It merely happens to find the value where the minimum occurs.

i too quite did it the same way..........took as x=y=z...& then did it your way

Arnav Das - 6 years, 1 month ago

You still need to explain what you mean by "maximize y and minimize the other two". For example, why not $y = 2, x = 1, z = 1$ ? that is a larger y, and a smaller x/z. As such, that logic is flawed.

You could approach this by first justifying why $z = x$ at the maximum, and then deal with this as one-variable calculus problem (where we substitute $y = \frac{ 2}{x^2}$ .

Calvin Lin Staff - 6 years, 1 month ago

Because each variable must be changed in a way that $x<y>z$ , but still meet the condition that $xyz=2$ . Squaring and square rooting is the way to achieve this effect. It will not work with any other exponent or rooting. Also, it said the minimum, and remember, minimizing is consistency.

For example:

$(2^{\frac{1}{3}})^\frac{1}{3}\times(2^{\frac{1}{3}})^{3} \times(2^{\frac{1}{3}})^\frac{1}{3}\ne2$

$(2^{\frac{1}{3}})^\frac{1}{4}\times(2^{\frac{1}{3}})^{4} \times(2^{\frac{1}{3}})^\frac{1}{4}\ne2$

Feathery Studio - 6 years, 1 month ago

What is wrong with $x = 1, y = 2 , z = 1$ which satisfies $xyz = 2$ and $x < y > z$ ?

Why must we be limited to "squaring and square rooting"?

What do you mean by "minimizing is consistency"? I don't understand the intention of your later 2 equations.

Calvin Lin Staff - 6 years, 1 month ago

@Calvin Lin – Yeah, the order in which I did things was messed up... the equations were part of my point that the only thing we can do is squaring and square rooting.

Proof that squaring and square rooting is the only way:

I must begin by saying that $y$ is inversely proportional to $xz$ , therefore, we perform opposite operations to either. For example, if we multiply $xz$ by $2$ , then we divide $y$ by $2$ .

$(2^{\frac{1}{3}})^\frac{1}{n}\times(2^{\frac{1}{3}})^{n} \times(2^{\frac{1}{3}})^\frac{1}{n} = 2^{1}$

$2^{\frac{1}{3n}}\times2^{\frac{n}{3}} \times2^{\frac{1}{3n}} = 2^{1}$

$\frac{1}{3n} + \frac{n}{3} + \frac{1}{3n} = \frac{2}{3n} + \frac{n}{3} = 1$

$2 + n^{2} = 3n$

$n^{2} - 3n + 2 = 0$

By quadratic formula,

$n = 1$ or $n = 2$

Because $1$ would be redundant and not get us anywhere, $n = 2$ .

Anyways, the reason that we square/square root instead of making $x=1$ , $y=2$ , and $z=1$ is that the change is much more slight than the latter scenario, therefore, the variables will be closer to each other in value and in effect, the expression will be minimized.

Feathery Studio - 6 years ago

Lu Chee Ket
Nov 11, 2015

10.607831003+ is a better proximity as the minimum value wanted.

x = 1.1969307793+
y = 1.3960209170+
z = 1.1969307793+

Their product is very close to 2.

There is no problem on writing up your own solution. But I can't get it, really made no sense. Sorry.

Reineir Duran - 5 years, 5 months ago

Why isn't it y^4?

The answer is 10.6.

2 solutions

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