Basic Math Problem of the Week 12/17/2017

[PF PotW Robot]

Here is this week's basic math problem. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Using spoiler tags is optional. Occasionally there will be prizes for extraordinary or clever methods.

Minimize ##\frac{2x^3+1}{4y(x-y)}## given ##x\ge -\frac{1}{2}## and ##\frac{x}{y}>1##

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[mfb]

Spoiler: significant help

Introduce c with y=cx

 

[Delta2]

This is interesting, wolfram(non Pro) can't solve it, it seems to crash reporting as global minimum a point where x=y (objective function is not defined/becomes infinite for x=y). Seems wolfram can't handle properly the infinities along the line y=x...

 

[Delta2]

Ok following @mfb suggestion it becomes ##min\frac{2x^3+1}{4x^2c(1-c)}, x\geq-0.5, 0<c<1##.

Now if we proceed by first minimizing for x we get ##\frac{\partial f(x,c)}{\partial x}=0\Rightarrow \frac{1}{2c(1-c)}-\frac{1}{2c(1-c)x^3}=0## or ##x=1##

and then minimizing for c ##\frac{\partial f(1,c)}{\partial c}=0\Rightarrow \frac{3}{4}\frac{1-2c}{(c(1-c))^2}=0\Rightarrow c=0.5##, so the minimum is at ##x=1,y=cx=0.5##

BUT is this procedure right (does it give us the real minimum of the original problem?) I mean first to minimize as to x and then minimize as to c?
Probably this procedure isn't right for general objective functions and constraints but seems to work for this problem...

 

[mfb]

A derivative of zero is not the only way something can be minimal.

The choice of c makes the expression factorize (f(x,c)=g(x)*h(c)), and if you are careful with the sign you can minimize both factors separately.

 

[StoneTemplePython]

my take:

based on constraints, we know x, y have the same sign. And the denominator is always positive. By inspection we see that even if ##-0.5 \leq x \lt 0##, the numerator is positive. We also know ##x \neq 0## (otherwise we can't satisfy ##\frac{x}{y} \gt 1##).

Thus we're trying to minimize a positive objective function.

- - - -
here's a sketch of a solution based on inequalities.

suppose we restrict our domain and consider only ##x \gt 0##

Spoiler

our objective function is minimized if and only if it's positive square root is, so we consider that easier problem instead. (Normally its the other way around!)
- - - -
goal minimize: ##\Big(\frac{2x^3+1}{4y(x-y)}\Big)^\frac{1}{2}##

##\text{ObjectiveFunction}^\frac{1}{2} = \Big(\frac{2x^3+1}{4y(x-y)}\Big)^\frac{1}{2} = \Big(3 \frac{\frac{1}{3}(2x^3+1)}{4y(x-y)}\Big)^\frac{1}{2}##
- - - -
by ##AM\geq GM##, we know

##\frac{1}{3}(2x^3+1) = \frac{1}{3}(x^3 + x^3+1^3) \geq \big(x^3x^31^3\big)^\frac{1}{3} = x^2##
with equality iff ##x^3 = x^3 =1 ##

i.e. the only way to make this inequality not strict is if ##x = 1## which is legal, given our constraints
(it pays to keep an eye on sharpness of these inequalities as we go)
- - - -

##\text{ObjectiveFunction}^\frac{1}{2} = \Big(3 \frac{\frac{1}{3}(2x^3+1)}{4y(x-y)}\Big)^\frac{1}{2} \geq \Big(3 \frac{x^2}{4y(x-y)}\Big)^\frac{1}{2} = \frac{\sqrt{3}}{2} \frac{ x}{y^\frac{1}{2}(x-y)^\frac{1}{2}}##
- - - -
now we'd like to maximize the value in the denominator, via applying ##AM \geq GM##:

##y^\frac{1}{2}(x-y)^\frac{1}{2} \geq \frac{1}{2}\big(y + (x - y)\big) = \frac{1}{2}x##

this inequality is an equality if and only if

##x - y = y##

i.e. ##x = 2y##.

Note this satisfies our constraint that ##\frac{x}{y} \gt 1##, and both ##y## and ##(x-y)## are positive, which satisfies conditions for ##AM \geq GM##
- - - -
this gives us

##\text{ObjectiveFunction}^\frac{1}{2} \geq \frac{\sqrt{3}}{2} \frac{ x}{y^\frac{1}{2}(x-y)^\frac{1}{2}} \geq \frac{\sqrt{3}}{2} \frac{ x}{\frac{1}{2}x} = 2\frac{\sqrt{3}}{2} \frac{ x}{x} = \sqrt{3} ##

thus we have

##\text{ObjectiveFunction}^\frac{1}{2} \geq \sqrt{3}##
aka

##\text{ObjectiveFunction} \geq 3##

with equality iff

##x = 1##

and

## 1 = x = 2y##

i.e. ## y =\frac{1}{2}##

- - - -

edit:
it seems that there is a matching score in bottom left quadrant.

to finish this off and address bottom left quadrant, with similar approach to the above:

Spoiler

taking advantage of positivity while working in the bottom left quadrant i.e. where ##0\gt y \gt x \geq -0.5## :

min ##\text{ObjectiveFunction}^\frac{1}{2} = \Big(\frac{2x^3+1}{4y(x-y)}\Big)^\frac{1}{2}= \Big(\frac{1}{4}\frac{2x^3+1}{y(x-y)}\Big)^\frac{1}{2} = \Big(\frac{1}{2}\frac{\big(2x^3+1\big)^\frac{1}{2}}{(-y)^\frac{1}{2}(y-x)^\frac{1}{2}}\Big) \geq \Big(\frac{1}{2}\frac{\big( 2x^3+1\big)^\frac{1}{2}}{\frac{1}{2} (-y + y - x)}\Big) = \Big(\frac{1}{2}\frac{\big(2x^3+1\big)^\frac{1}{2}}{\frac{1}{2} (-x)}\Big)##

with equality conditions of

##-y = y - x##

i.e. ##2y = x##

and confirming that ##-y \gt 0## and ##(y-x) \gt 0##

while messy, the above is now a single variable problem that can be simplified and solved

i.e. (recalling our domain restrictions) we could also square the above

and look at minimizing

##g(x) = \frac{ 2x^3 + 1}{x^2} ##

which we know is always positive (because the objective function is), and if we differentiate it once, we see ##g'(x) = 2 - \frac{2}{x^3} \gt 0## for ##x \lt 0##, i.e. the curve slopes up over our domain. (We could also note that the positive solution evaluates to 0 when x = 1.)

If we differentiate it twice, we see that ##g''(x) = \frac{6}{x^4} \gt 0## for any ##x \neq 0## confirming convexity. This means the upward sloping tangent line at ##x = -0.5## is a lower bound on the function in our domain i.e. we can only increase the value of the function as x increases / goes to the right of ##-0.5##. We could perhaps decrease the value ##g(x)## by going to the left, but we have a constraint that ##x \geq -0.5##. This explains the other way of achieving the minimum value of 3, in the bottom left quadrant with ##f(-0.5, -0.25)##

 

[PetSounds]

Here's my go at it. I haven't read any of the other solutions, and given that I only know high school math, I doubt they'd do me much good anyway.

Because of the constraints on ##x##, we know the numerator is greater than zero. Because the absolute value of ##x## is greater than the absolute value of ##y##, and ##x/y > 1##, we know that the denominator also is greater than zero.

Consider only the denominator. For any value of ##x##, the value of ##y## that maximizes ##4xy - 4y^2## is ##y = x/2##. (Treat the denominator as a quadratic with ##x## as a constant.) Sine ##y## only appears in the denominator, this is the value of ##y## which will minimize the function.

We then have the fraction as a function of ##x##.

##f(x) = \frac {2x^3 + 1}{x^2}
= 2x + \frac {1}{x^2}##

Minimizing this function gives us a value of ##x = 1##. Thus ##y = 1/2##.

##\frac {2x^3 + 1}{4y(x-y)} = 3##

The minimum is 3.

EDIT: Hey! Looks like I got it right using only high school calculus.

 

[mfb]

Minimizing this function gives us a value of ##x=1##

While that is a minimum, it is not the only one.

 

[PetSounds]

While that is a minimum, it is not the only one.

Right. Because of the restriction on ##x##, there's also a minimum at ##x = -1/2##. Which means ##y = -1/4## and, again, the original fraction is equal to 3.

 

[Delta2]

A derivative of zero is not the only way something can be minimal.

The choice of c makes the expression factorize (f(x,c)=g(x)*h(c)), and if you are careful with the sign you can minimize both factors separately.

That factorization might explain why minimizing first with respect to x and then with respect to c works. But anyway I think the procedure of successive minimization is correct as long as each minimum exists. (@PetSounds also in post #7 does successive minimization , first minimizes wrt y and then wrt x). But it might be the case (depending on the form of the objective function and the constraints) that these minimums/maximums don't exist, however the "combined" extreme exists.

 

[PetSounds]

That factorization might explain why minimizing first with respect to x and then with respect to c works. But anyway I think the procedure of successive minimization is correct as long as each minimum exists. (@PetSounds also in post #7 does successive minimization , first minimizes wrt y and then wrt x). But it might be the case (depending on the form of the objective function and the constraints) that these minimums/maximums don't exist, however the "combined" extreme exists.

I'm still in high school, so forgive me if I'm repeating or misunderstanding you---a lot of this goes over my head. The reason the successive minimization works in this case is that the original fraction is always positive and ##y## only appears in the denominator. Thus we can simply find the optimum value of ##y## (in terms of ##x##) by maximizing the denominator, without regard for the numerator. If ##y## appeared in the numerator, or if the constraints on the variables allowed for both positive and negative values of the fraction, my method wouldn't work (at least, not as cleanly).

 

[Delta2]

I'm still in high school, so forgive me if I'm repeating or misunderstanding you---a lot of this goes over my head. The reason the successive minimization works in this case is that the original fraction is always positive and ##y## only appears in the denominator. Thus we can simply find the optimum value of ##y## (in terms of ##x##) by maximizing the denominator, without regard for the numerator. If ##y## appeared in the numerator, or if the constraints on the variables allowed for both positive and negative values of the fraction, my method wouldn't work (at least, not as cleanly).

Don't worry you are not repeating me, your ideas in post #7 were original, I just think (cant find a proof yet) that successively minimizing gives correct solutions (regardless if the y was appearing in the numerator or if the fraction had positive and negative values and all that) as long as we can correctly find the separate minimums and as long these minimums exist. That's all. ( I ll just have to find and look at my course notes on mathematical programming long ago (20 years) when I was student at university...)