Basic Math Problem of the Week 1/02/2017

[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.

Let $a,\,b$ be non-negative numbers with $\sqrt{1-\frac{a^2}{4}}+\sqrt{1-\frac{b^2}{16}}=\frac{3}{2}$

Find the maximum value of $ab$

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

 

[Charles Link]

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.

Let $a,\,b$ be non-negative numbers with $\sqrt{1-\frac{a^2}{4}}+\sqrt{1-\frac{b^2}{16}}=\frac{3}{2}$

Find the maximum value of $ab$

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

A standard calculation using Lagrangian multipliers yielded the result that, at the maximum, $a=1.20$ and $b=2.85$ (approximately). I found the roots by hand without the benefit of a computer or calculator, so I could be off on the second decimal place. That would make the maximum product $ab=3.42$.

 

[fresh_42]

I think this is a bit too low.

 

[.Scott]

I have a=1.322876, b=2.645751, ab=3.5
That's from Excel Spreadsheet

 

[Charles Link]

I have a=1.322876, b=2.645751, ab=3.5
That's from Excel Spreadsheet

It looks reasonable. I could easily have been off that much in my hand calculation for $a$ and $b$. Algebraically I had a clumsy (but accurate) expression involving $a$ and was plugging in different values of $a$ to make the two sides of the equation approximately equal. $\\$ Editing: My expression for $a$ from the Lagrangian multiplier processing is $(9-2a^2)(\sqrt{1-\frac{a^2}{4}})+(\frac{9}{2})a^2=12$. Yes, I tried @.Scott 's $a=1.32$ and it fits it almost perfectly.

 

[fresh_42]

This is correct (I think), although a bit of an usual way to write it. A bit engineerish

 

[.Scott]

I had my spread sheet set up to handle a guess for A and B then it would compute how far from the that 1.5 value those guesses got you.
From there, it did 6 iterations of an approximate Newton-Raphson (I supplied a guess at the first derivative).
When the 1st derivative was close enough, I would see it close in just a couple of iterations.
Then I took the result from the last iteration and computed the a*b product.

So I would pick an a, adjust my initial b and derivative to get it to close, and note how far my a*b changed.
Once I started getting close, I didn't need to change the derivative or the initial guess any more - just the a.

I agree - very Engineeringish. I am a Software Engineer.

I also notice that a*a = 1.75 and b*b = 7.

 

[fresh_42]

The trick is to find an argument, why $b=2a$ has to hold.

 

[Charles Link]

The trick is to find an argument, why $b=2a$ has to hold.

So $a=\frac{\sqrt{7}}{2}$ and $b=\sqrt{7}$. See my also my edited post 5. I can plug in this value for $a$ and show it is exact. See post 12 where I took the expression of post 5 and converted to a cubic equation in $a^2$.

 

[fresh_42]

Both roots are strictly monotone on their domains. So the maximum of $ab$ is an extremum of each root (under the given sum). And the maximum of $ab$ given $a+b=c$ leads to $b=2a$. However, I haven't found a more mathematical way to express this, i.e. why it is allowed to require $a+b=c$ instead of $f(a)+g(b)=c$. I assume monotony as the key here.

 

[Buzz Bloom]

My spreadsheet effort led to the following:
a = 1.322875622

ab = 3.49999999999999​

a = 1.322875623 through 1.322875688

ab = 3.5​

a = 1.322875689

ab = 3.49999999999999
​

 

[Charles Link]

See my post 5 at the bottom. The result after some algebra is $4a^6+29 a^4-207a^2+252=0$. This is a cubic in $a^2$. Factoring this gives $(4u-7)(u^2+9u-36)=0$ where $u=a^2$, and it is evident $u=a^2=\frac{7}{4}$ is one solution. The corresponding $b^2=7$, with the result that $ab=\frac{7}{2}$ is the maximum value for the product $ab$. $\\$ Having @.Scott 's spreadsheet solution was very helpful,( thank you @.Scott ), in solving the for the solution. Without knowing there was an exact solution, I probably would not have had any reason to look for one. :) $\\$ Additional item: The quadratic factors to $(u-3)(u+12)=0$. One solution is $u=3$.This means $a=\sqrt{3}$ should also be a local extremum of some kind. $\\$ Editing: For the case of $a=\sqrt{3}$, $b=0$, so it is not a maximum.

 

[mfb]

Make it symmetric.

A maximum of ab is also a maximum of ab/8=xy where x=a/2 and y=b/4. Now we have the additional condition $\sqrt{1-x^2}+\sqrt{1-y^2}=\frac 3 2$
It is quite easy to see that xy is maximized when x=y, or $1-x^2=1-y^2=\frac {9} {16}$, which leads to $x=y=\frac{\sqrt{7}}{4}$ and therefore $a=\frac{\sqrt{7}}{2}$, $b=\sqrt{7}$

Edit: Fixed some denominators.

 

[Charles Link]

Make it symmetric.

A maximum of ab is also a maximum of ab/8=xy where x=a/2 and y=b/4. Now we have the additional condition $\sqrt{1-x^2}+\sqrt{1-y^2}=\frac 3 2$
It is quite easy to see that xy is maximized when x=y, or $1-x^2=1-y^2=\frac {9} {16}$, which leads to $x=y=\frac{\sqrt{7}}{2}$ and therefore $a=\sqrt{7}$, $b=\frac{\sqrt{7}}{2}$

A very minor correction: $x=y=\frac{\sqrt{7}}{4}$, so that $a=\frac{\sqrt{7}}{2}$, and $b=\sqrt{7}$, but this is incidental. The method is clearly much simpler than applying Lagrangian multipliers to the original equation.

 

[StoneTemplePython]

Make it symmetric.

A maximum of ab is also a maximum of ab/8=xy where x=a/2 and y=b/4. Now we have the additional condition $\sqrt{1-x^2}+\sqrt{1-y^2}=\frac 3 2$
It is quite easy to see that xy is maximized when x=y, or $1-x^2=1-y^2=\frac {9} {16}$, which leads to $x=y=\frac{\sqrt{7}}{2}$ and therefore $a=\sqrt{7}$, $b=\frac{\sqrt{7}}{2}$

I like this restatement -- it cleans things up nicely.

I'd probably run the argument in full with convexity and $GM \leq AM$ as below.

- - - -

The map $f(u) = \sqrt{1-u^2}$ is strictly negative convex (over real solutions) as confirmed by second derivative test.

Hence we have

$\frac{3}{4} = \frac{1}{2}\sqrt{1-x^2} +\frac{1}{2}\sqrt{1-y^2} = E\big[f(u)\big] \leq f(E\big[u\big]) = \sqrt{\frac{1}{2}(1-x^2) + \frac{1}{2}(1-y^2) } = \sqrt{1-\big(\frac{x^2 + y^2}{2}\big)}$

with equality iff $x = y$ by strict convexity.

- - - - - -
edit:
To make this coherent it's better to for me to say I'm using the mapping

$f(u)=\sqrt{u}$

not $f(u) = \sqrt{1-u^2}$

as pointed out by @Charles Link

This function is is still strictly negative convex, and gives us the above inequality.

In the interest of clarity, it should say

"with equality iff $(1-x^2) = (1-y^2)$ by strict negative convexity. This is equivalent to $x^2 = y^2$. Since our domain is confined to real non-negative $x, y$ this is equivalent to insisting on $x = y \geq 0$"
- - - - - -

Taking advantage of positivity, square both sides.

$\frac{9}{16} \leq 1-\big(\frac{x^2 + y^2}{2}\big)$

$- \frac{7}{16} \leq -\big(\frac{x^2 + y^2}{2}\big)$

multiply by negative one

$\frac{x^2 + y^2}{2} \leq \frac{7}{16}$

recognizing that our actual objective function is maximized iff $xy$ is maximized, we see

$xy = \big(x^2y^2\big)^\frac{1}{2} \leq \frac{x^2 + y^2}{2} \leq \frac{7}{16}$

By $GM \leq AM$ and the the above inequality. Our function reaches maximum iff x = y.

 

[Charles Link]

I like this restatement -- it cleans things up nicely.

I'd probably run the argument in full with convexity and $GM \leq AM$ as below.

- - - -

The map $f(u) = \sqrt{1-u^2}$ is strictly negative convex (over real solutions) as confirmed by second derivative test.

Hence we have

$\frac{3}{4} = \frac{1}{2}\sqrt{1-x^2} +\frac{1}{2}\sqrt{1-y^2} = E\big[f(u)\big] \leq f(E\big[u\big]) = \sqrt{\frac{1}{2}(1-x^2) + \frac{1}{2}(1-y^2) } = \sqrt{1-\big(\frac{x^2 + y^2}{2}\big)}$

with equality iff $x = y$ by strict convexity.

taking advantage of positivity, square both sides.

$\frac{9}{16} \leq 1-\big(\frac{x^2 + y^2}{2}\big)$

$- \frac{7}{16} \leq -\big(\frac{x^2 + y^2}{2}\big)$

multiply by negative one

$\frac{x^2 + y^2}{2} \leq \frac{7}{16}$

recognizing that our actual objective function is maximized iff $xy$ is maximized, we see

$xy = \big(x^2y^2\big)^\frac{1}{2} \leq \frac{x^2 + y^2}{2} \leq \frac{7}{16}$

By $GM \leq AM$ and the the above inequality. Our function reaches maximum iff x = y.

@StoneTemplePython My formal mathematics might be lacking somewhat here, but I have a question or two that might make it easier to follow in detail. If I am not mistaken you are applying Jensen's inequality, and isn't the function for this case $f(u)=\sqrt{u}$ ? Also, isn't $x_1=1-x^2$ and $x_2=1-y^2$, with weights of 1/2 for each? The condition $x_1=x_2$ would be equivalent to $x=y$. $\\$ In any case, I found your solution quite interesting. :)

 

[StoneTemplePython]

@StoneTemplePython My formal mathematics might be lacking somewhat here, but I have a question or two that might make it easier to follow in detail. If I am not mistaken you are applying Jensen's inequality

Strictly speaking you can call it Jensen's Inequality-- the general statement that

$E\big[f(u)\big] \leq f(E\big[u\big])$ is Jensen's Inequality for a negative convex function. However

$f(p u_1 + (1-p)u_2) \leq (p)f(u_1) + (1-p)f(u_2)$ for $p \in [0,1]$

is actually the definition of convexity for $u_1, u_2$ in a suitable domain. For negative convexity, the inequality sign flips to:

$f(p u_1 + (1-p)u_2) \geq (p)f(u_1) + (1-p)f(u_2)$ for $p \in [0,1]$

in our problem, set $p := \frac{1}{2}$

So either Jensen's Inequality or the definition of negative convexity works when you're just doing an average of two terms.

and isn't the function for this case $f(u)=\sqrt{u}$ ? Also, isn't $x_1=1-x^2$ and $x_2=1-y^2$, with weights of 1/2 for each? The condition $x_1=x_2$ would be equivalent to $x=y$. $\\$ In any case, I found your solution quite interesting. :)

actually I was thinking of a few different mappings-- and kind of used a motley crew of them

I like your idea better and will update my post.

 

[Charles Link]

It is still interesting that the method of Lagrangian multipliers furnished the same result. (See posts 5,9 and 12 above). To give the some of the details of that solution, the function $f(x,y, \lambda)=xy+\lambda(\sqrt{1-\frac{x^2}{4}}+\sqrt{1-\frac{y^2}{16}}-\frac{3}{2})$ is examined for maxima by solving the three equations $\frac{\partial{f}}{\partial{x}}=0$, $\frac{\partial{f}}{\partial{y}}=0$, and $\frac{\partial{f}}{\partial{\lambda}}=0$. $\\$ The last equation is simply the constraint equation. (Note $x=a$, and $y=b$). This results in the equation of post 5 for $a$: $(9-2a^2)\sqrt{1-\frac{a^2}{4}}+(\frac{9}{2})a^2=12$. After some algebra, this results in the cubic equation in $a^2$ given in post 12 : $4a^6+29a^4-207a^2+252=0$. This factors as $(4u-7)(u+12)(u-3)=0$, and gives $u= a^2=\frac{7}{4}$ as the root of interest.

 

[StoneTemplePython]

edit -- I had a follow up idea that had a fatal bug. Unable to delete posting unfortunately.

 

[StoneTemplePython]

Here is the geometric interpretation of the problem that I was looking for.

Consider Minkowski's Light Cone and the Lorentz Product. It gives something that looks like backward Cauchy Schwarz:

$[\mathbf x, \mathbf x]^\frac{1}{2}[\mathbf y, \mathbf y]^\frac{1}{2}\leq [\mathbf x, \mathbf y]$

this may be proven several ways, including via $GM \leq AM$. If you go through the proof, you can get a slightly more detailed inequality that:

$\big([\mathbf x, \mathbf x][\mathbf y, \mathbf y]\big)^\frac{1}{2}\leq \frac{1}{2} \big([\mathbf x, \mathbf x] + [\mathbf y, \mathbf y]\big) \leq [\mathbf x, \mathbf y]$

the inequality is strict except in a special case -- said case for our problem implies that $x_1 = \frac{a}{2} = \frac{b}{4} = y_1$

- - - -
our original problem constraint can now be written in terms of Lorentz products

$[\mathbf x, \mathbf x]^\frac{1}{2} + [\mathbf y, \mathbf y]^\frac{1}{2} = \sqrt{1-\frac{a^2}{4}}+\sqrt{1-\frac{b^2}{16}}=\frac{3}{2}$

taking advantage of non-negativity, we may square both sides.

$[\mathbf x, \mathbf x] + [\mathbf y, \mathbf y] + 2\big([\mathbf x, \mathbf x][\mathbf y, \mathbf y]\big)^\frac{1}{2} =\frac{9}{4}$

for convenience, divide each side by two and get

$\frac{9}{8} = \frac{1}{2} \Big([\mathbf x, \mathbf x] + [\mathbf y, \mathbf y]\Big) + \Big(\big([\mathbf x, \mathbf x][\mathbf y, \mathbf y]\big)^\frac{1}{2}\Big) \leq \Big([\mathbf x, \mathbf y]\Big) + \Big([\mathbf x, \mathbf y]\Big) = 2[\mathbf x, \mathbf y]$

dividing by 2 gives us

$\frac{9}{16} \leq [\mathbf x, \mathbf y] = 1 - \big(x_1\big)\big(y_1\big) = 1 - \big(\frac{a}{2}\big)\big(\frac{b}{4}\big) = 1 - \frac{ab}{8}$

subtract one from each side

$\frac{-7}{16} \leq -\frac{ab}{8}$

multiply each side by $-8$

$\frac{7}{2} \geq ab$

and again this inequality is strict, unless $a = \frac{b}{2}$

 

[PetSounds]

Are these done with?

 

[Greg Bernhardt]

I will reboot the PotW feature soon! There was a little mishap that I need to fix.