What is the maximum value of $a^2b$?

[anemone]

Let $a$ and $b$ be positive real numbers such that $a+b+\sqrt{2a^2+2ab+3b^2}=4$.

Determine the maximum value of $a^2b$.

 

[topsquark]

Let $a$ and $b$ be positive real numbers such that $a+b+\sqrt{2a^2+2ab+3b^2}=4$.

Determine the maximum value of $a^2b$.

By inspection it's 8. It's always 8 when you have no idea...

-Dan

 

[MarkFL]

My solution:

Spoiler

We have the objective function:

\(\displaystyle f(a,b)=a^2b\)

Subject to the constraint:

\(\displaystyle g(a,b)=a+b+\sqrt{2a^2+2ab+3b^2}-4=0\)

Using Lagrange Multipliers, we obtain the system:

\(\displaystyle 2ab=\lambda\left(1+\frac{2a+b}{\sqrt{2a^2+2ab+3b^2}}\right)\)

\(\displaystyle a^2=\lambda\left(1+\frac{a+3b}{\sqrt{2a^2+2ab+3b^2}}\right)\)

This implies:

\(\displaystyle \frac{2b}{2a+b+\sqrt{2a^2+2ab+3b^2}}=\frac{a}{a+3b+\sqrt{2a^2+2ab+3b^2}}\)

Cross multiply and combine like terms:

\(\displaystyle ab+6b^2+2b\sqrt{2a^2+2ab+3b^2}=2a^2+a\sqrt{2a^2+2ab+3b^2}\)

Get radicals on same side:

\(\displaystyle -2a^2+ab+6b^2=a\sqrt{2a^2+2ab+3b^2}-2b\sqrt{2a^2+2ab+3b^2}\)

Factor:

\(\displaystyle -(a-2b)(2a+3b)=(a-2b)\sqrt{2a^2+2ab+3b^2}\)

Since both $a$ and $b$ are positive, we cannot have:

\(\displaystyle -(2a+3b)=\sqrt{2a^2+2ab+3b^2}\)

And so our only critical point will come from:

\(\displaystyle a=2b\)

Substituting into the constraint, we obtain:

\(\displaystyle 2b+b+\sqrt{2(2b)^2+2(2b)b+3b^2}-4=0\)

\(\displaystyle 3b+b\sqrt{15}=4\)

\(\displaystyle b=\frac{4}{3+\sqrt{15}}\cdot\frac{3-\sqrt{15}}{3-\sqrt{15}}=\frac{2}{3}(\sqrt{15}-3)\implies a=\frac{4}{3}(\sqrt{15}-3)\)

We find:

\(\displaystyle f\left(\frac{4}{3}(\sqrt{15}-3),\frac{2}{3}(\sqrt{15}-3)\right)=\frac{32}{27}(\sqrt{15}-3)^3=\frac{64}{9}(7\sqrt{15}-27)\)

To verify this is a maximum, let's pick another point on the constraint, such as:

\(\displaystyle (a,b)=\left(1,\frac{1}{2}(\sqrt{30}-4)\right)\)

We find:

\(\displaystyle f\left(1,\frac{1}{2}(\sqrt{30}-4)\right)=\frac{1}{2}(\sqrt{30}-4)<\frac{64}{9}(7\sqrt{15}-27)\)

And so we may conclude that on the given domain, we have:

\(\displaystyle f_{\max}=\frac{64}{9}(7\sqrt{15}-27)\)

 

[Albert1]

Let $a$ and $b$ be positive real numbers such that $a+b+\sqrt{2a^2+2ab+3b^2}=4---(1)$.

Determine the maximum value of $a^2b---(2)$.

my solution:

Spoiler

using $AP\geq GP$
from (2) we have $\dfrac {a^4+b^2}{2}\geq a^2b$ equality occurs when $a^4=b^2$ or $b=a^2---(3)$
put (3) to (1)we obtain :$a+a^2+\sqrt {2a^2+2a^3+3a^4}=4$ or
$\sqrt {2a^2+2a^3+3a^4}=4-a-a^2---(4)$
square both sides of (4) and rearrange we get:
$2a^4+9a^2+8a-16=0---(5)$
let $f(a)=2a^4+9a^2+8a-16$
for $f(0.9)<0, f(1)=2+9+8-16=3>0$
the solution of (5) : $0.9<a<1$
using Newton's Method:$x_{n+1}=x_n-\dfrac {f(x_n)}{f'(x_n)}$
take $x_0=1,$ we have $x_1=1-\dfrac {3}{34}\approx 0.911765$
continue this process the only positive solution of (5):$a\approx 0.9066$
and $max(a^2b)=a^4=(0.9066)^4\approx 0.68$ seemed smaller than MarkFL's answer

 

[Opalg]

[sp]This is just a variation on Mark's solution, from a geometric viewpoint (so as to avoid Lagrange multipliers at all costs (Tongueout)).

Write the equation as $\sqrt{2a^2 + 2ab + 3b^2} = 4 - a - b$ and square both sides, to get $2a^2 + 2ab + 3b^2 = 16 - 8a - 8b + a^2 + b^2 + 2ab$, which simplifies to $$a^2 + 2b^2 + 8a + 8b = 16.$$ I prefer to use $x$ and $y$ instead of $a$ and $b$, so this becomes $$x^2 + 2y^2 + 8x + 8y = 16.$$ (This is the equation of an ellipse centred at $(-4,-2)$.) We want to find the maximum value of $x^2y$ at a point on this ellipse. So we need to find the largest value of $k$ for which the curve $x^2y=k$ intersects the ellipse somewhere in the positive quadrant. That will happen when the two curves just touch at a single point. In that case, the two curves will share a common tangent at that point, and in particular they will have the same slope at that point. So, differentiate the equations of the curves to get $$2x + 4yy' + 8 + 8y' = 0,\qquad 2xy + x^2y' = 0.$$ If those equations both hold at the same point $(x,y)$ then $$y' = -\frac{x+4}{2(y+2)} = -\frac{2y}x,$$ so that $x(x+4) = 4y(y+2)$. Write that as $x^2 - 4y^2 + 4(x-2y) = 0.$ Factorise the difference of two squares, getting $(x-2y)(x+2y+4) = 0.$

Now, just as in Mark's solution, $x+2y+4$ cannot be zero, so we must have $x=2y.$ Substitute that into the ellipse equation, getting $6y^2 + 24y - 16 = 0$, or $3y^2 + 12y -8=0.$ Solve that quadratic, to get the positive solution $y = -2 + \frac23\sqrt{15}.$

From the quadratic, $y^2 = -4y + \frac83$,so $$y^3 = -4y^2 + \tfrac83y = -4\bigl(-4y + \tfrac83\bigr) + \tfrac83y = \tfrac{56}3y - \tfrac{32}3 = \tfrac{56}3\bigl(-2 + \tfrac23\sqrt{15}\bigr) - \tfrac{32}3 = -48 + \tfrac{112}9\sqrt{15}.$$

Finally, \(\displaystyle x^2y = (2y)^2y = 4y^3 = -192 + \tfrac{448}9\sqrt{15} = 64\bigl(-3 + \tfrac79\sqrt{15}\bigr).\) That represents the maximum value of $x^2y$ on the ellipse. Mercifully, this agrees with Mark's result.

In the graph below, the orange ellipse is $x^2 + 2y^2 + 8x + 8x = 16,$ the blue curve is $x^2y = 64\bigl(-3 + \frac79\sqrt{15}\bigr)$, and the green line is $x=2y.$

[graph]zdzkxhk1dn[/graph]​

[/sp]

 

[Albert1]

according to MarkFL and Opalg 's solution :
$max(a^2b)\approx 0.79$ (occurs at $b=\dfrac {a}{2}$)
and my solution using $AP\geq GP$ and Newton's Method
$max(a^2b)\approx 0.68$ (occurs at $ b=a^2$)
why? (I am not convinced)
if we want to find $max(x)\leq y, (for \,\ x,y>0)$
$y$ should be as smaller as possible,right ?
$a^2b\leq 0.68$ , so $a^2b<0.79 $ must be true

 

[anemone]

Hi all,

Sorry for the late reply! But I can explain...(Smile) I felt under the weather since last Sunday night and by last Thursday, I felt worst and so I went to see a doctor and took the medical leave on Friday and went back from Singapore to Malaysia on Friday and back to Singapore again yesterday...I didn't have the time to check MHB out until now...I am still feeling a bit under the weather today though...

Okay, I am done talking about myself, now it is high time to give merit to MarkFL and Opalg for their beautiful solution and explanation! Bravo!(Cool)

 

[Albert1]

Thanks for your answer.
I hope you have recovered from your illness by now

 

[Opalg]

Sorry to hear you've been unwell, anemone. Hope you're fully recovered soon.

 

[juantheron]

Spoiler

Given $a+b+\sqrt{2a^2+2ab+3b^2} = 4$ Here $a,b>0$

Now Using $\bf{Cauchy\; Schwarz}$ Inequality

$\displaystyle (a^2+2b^2)(2+1)\geq 2(a+b)^2\Rightarrow (a^2+2b^2)\geq \frac{2}{3}(a+b)^2$

Now $\displaystyle 2a^2+2ab+3b^2=(a+b)^2+(a^2+2b^2)\geq (a+b)^2+\frac{2}{3}(a+b)^2$

So $\displaystyle \sqrt{2a^2+2ab+3b^2}\geq \sqrt{\frac{5}{3}}\sqrt{(a+b)^2} = \sqrt{\frac{5}{3}}(a+b)\;$

So $a+b+\sqrt{2a^2+2ab+3b^2}\geq (a+b)+\sqrt{\frac{5}{3}}(a+b)$

So $\displaystyle 4\geq (a+b)\left(1+\sqrt{\frac{5}{3}}\right).$

So $\displaystyle (a+b)\leq \frac{4\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

Now Using $\bf{A.M\geq G.M}$

So we get $\displaystyle \left(\frac{a}{2}\cdot \frac{a}{2}\cdot b\right)^{\frac{1}{3}}\leq \frac{(a+b)}{3}\leq \frac{4\sqrt{3}}{3(\sqrt{5}+\sqrt{3})} = \frac{4}{3+\sqrt{15}}$

So $\displaystyle a^2b\leq \frac{4\cdot 4^3}{(3+\sqrt{15})^3} = \frac{256}{(3+\sqrt{15})^3}$ and equality hold when $a=b$

 

[anemone]

Thanks for your answer.
I hope you have recovered from your illness by now

Hi Albert,

I am glad you liked my explanation in the PM and I hope it clears up your confusion and thanks for your concern!:)

Sorry to hear you've been unwell, anemone. Hope you're fully recovered soon.

Thanks Opalg for your best wish! :)

Spoiler

Given $a+b+\sqrt{2a^2+2ab+3b^2} = 4$ Here $a,b>0$

Now Using $\bf{Cauchy\; Schwarz}$ Inequality

$\displaystyle (a^2+2b^2)(2+1)\geq 2(a+b)^2\Rightarrow (a^2+2b^2)\geq \frac{2}{3}(a+b)^2$

Now $\displaystyle 2a^2+2ab+3b^2=(a+b)^2+(a^2+2b^2)\geq (a+b)^2+\frac{2}{3}(a+b)^2$

So $\displaystyle \sqrt{2a^2+2ab+3b^2}\geq \sqrt{\frac{5}{3}}\sqrt{(a+b)^2} = \sqrt{\frac{5}{3}}(a+b)\;$

So $a+b+\sqrt{2a^2+2ab+3b^2}\geq (a+b)+\sqrt{\frac{5}{3}}(a+b)$

So $\displaystyle 4\geq (a+b)\left(1+\sqrt{\frac{5}{3}}\right).$

So $\displaystyle (a+b)\leq \frac{4\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

Now Using $\bf{A.M\geq G.M}$

So we get $\displaystyle \left(\frac{a}{2}\cdot \frac{a}{2}\cdot b\right)^{\frac{1}{3}}\leq \frac{(a+b)}{3}\leq \frac{4\sqrt{3}}{3(\sqrt{5}+\sqrt{3})} = \frac{4}{3+\sqrt{15}}$

So $\displaystyle a^2b\leq \frac{4\cdot 4^3}{(3+\sqrt{15})^3} = \frac{256}{(3+\sqrt{15})^3}$ and equality hold when $a=b$

Very well done, jacks! But ...

Spoiler

Note that equality doesn't hold when $a=b$, your both statements

(i) $\displaystyle (a^2+2b^2)(2+1)\geq 2(a+b)^2\Rightarrow (a^2+2b^2)\geq \frac{2}{3}(a+b)^2$ and

(ii) $\displaystyle \left(\frac{a}{2}\cdot \frac{a}{2}\cdot b\right)^{\frac{1}{3}}\leq \frac{(a+b)}{3}$

suggest that equality occurs when $a=2b$.(Wink)