Find the greatest and smallest value of x + y

[anemone]

Let \(\displaystyle x,\;y\) be real numbers satisfying the condition:

\(\displaystyle x-3\sqrt{x+1}=3\sqrt{y+2}-y\)

Find the greatest value and the smallest value of \(\displaystyle P=x+y\).

 

[chisigma]

Let \(\displaystyle x,\;y\) be real numbers satisfying the condition:

\(\displaystyle x-3\sqrt{x+1}=3\sqrt{y+2}-y\)

Find the greatest value and the smallest value of \(\displaystyle P=x+y\).

Is...

$$P= x + y = 3\ (\sqrt{x+1} + \sqrt{y+2})\ (1)$$

... so that You can find the maximum and minimum of P using standard approach...

Kind regards

$\chi$ $\sigma$

 

[Klaas van Aarsen]

The standard approach is the method of Lagrange multipliers.

We consider the function
$$\Lambda(x,y,\lambda)= x + y + \lambda \big( (x - 3\sqrt{x+1}) - (3\sqrt{y+2} - y) \big)$$

To find the extrema of $(x+y)$ we set each of the partial derivatives to zero and solve.
Furthermore, we consider any boundary extrema.

Spoiler

This yields one extremum with value $9 + 3 \sqrt{15} \approx 20.62$.

Furthermore, we have 2 boundary extrema at $x=-1$ and $y=-2$.
Solving for those we find 2 more extrema that turn out to have the same value $4\frac 1 2 + \frac 3 2 \sqrt{21} \approx 11.37$.

In other words, the minimal value of P is $4\frac 1 2 + \frac 3 2 \sqrt{21} \approx 11.37$.
And the maximal value of P is $9 + 3 \sqrt{15} \approx 20.62$.

 

[anemone]

Hi MHB,

I once solved this problem using an algebraic method, but once I saw the reply from I like Serena, who tackled it using Lagrange Multipliers(I got the feeling that I like Serena solved it quite easily and there might be something "wrong" with this problem in terms of its difficulty), I then looked at my algebraic approach and shook my head relentlessly because I felt this problem might not be considered a challenge itself, and it would be a shame for me to post some trivial problem here, and so I threw the paper containing my solution into the dustbin right away...

Now, when I look back and decide I want to to post my algebraic solution, I find that I am stuck and don't know how to proceed! I am so frustrated by this, but this does not bring the inspiration back, no matter how hard that I cudgel my brain...

If you happen to know how to solve it using an algebraic approach, could you please show me the right way to do it? Thanks in advance...

 

[Klaas van Aarsen]

Hi MHB,

I once solved this problem using an algebraic method, but once I saw the reply from I like Serena, who tackled it using Lagrange Multipliers(I got the feeling that I like Serena solved it quite easily and there might be something "wrong" with this problem in terms of its difficulty), I then looked at my algebraic approach and shook my head relentlessly because I felt this problem might not be considered a challenge itself, and it would be a shame for me to post some trivial problem here, and so I threw the paper containing my solution into the dustbin right away...

Now, when I look back and decide I want to to post my algebraic solution, I find that I am stuck and don't know how to proceed! I am so frustrated by this, but this does not bring the inspiration back, no matter how hard that I cudgel my brain...

If you happen to know how to solve it using an algebraic approach, could you please show me the right way to do it? Thanks in advance...

Let \(\displaystyle x,\;y\) be real numbers satisfying the condition:

\(\displaystyle x-3\sqrt{x+1}=3\sqrt{y+2}-y\)

Find the greatest value and the smallest value of \(\displaystyle P=x+y\).

All right.
Let me give it a try.
Hope you like my algebraic solution.

Substitute $u=\sqrt{x+1}, \quad v=\sqrt{y+2} \qquad (1)$

Then we need to find the extremes of
$$P = u^2 + v^2 - 3$$
with the condition that:
$$(u^2-1) - 3u = 3v - (v^2 - 2)$$
$$(u^2+v^2) - 3(u+v) - 3 = 0$$

Now substitute \(\displaystyle u=r\cos\phi, \quad v=r\sin\phi \qquad (2)\)
That gives you:
$$P = r^2 - 3 \qquad\qquad\qquad (3)$$
$$r^2 - 3(r\cos\phi + r\sin\phi) - 3 = 0$$
$$r^2 - 3r\sqrt 2(\sin(\phi + \pi/4)) - 3 = 0$$

Define \(\displaystyle B = 3\sqrt 2(\sin(\phi + \pi/4)) \qquad (4)\)
Then we have:
$$r^2 - Br - 3 = 0$$
$$r_\pm = \frac 1 2(B \pm \sqrt{B^2 + 12}) \qquad (5)$$

From (3) we can see that $P$ takes its maximal value if $r$ takes its maximal value as well.
From (5) it follows that $r$ takes its maximal value for the + solution $r_+$ where $B$ has its maximal value.
And from (4) we can see that $B$ takes its maximal value when $\sin(\phi + \pi/4)=1$, so $B_{max} = 3\sqrt 2$.

Substituting $B_{max}$ back into $r_+$ with (5), and in turn into $P$ through (3), gives the maximum value for $P$.
Similarly we can find the lowest value for $P$.

 

[Klaas van Aarsen]

I'm thinking it's sort of your turn now.
Can you work out the solution using the method of the Lagrange Multipliers?
I skipped a couple of steps in that solution...

 

[anemone]

Hello I like Serena,

Thanks for providing me such a great solution to tackle the problem using trigonometric approach and I appreciate it and I also like it!

Your patience and willingness to answer my "request" touched me very much. And I know I approached it differently than you (because I didn't introduce another variable ($\theta$ in your case) in my attempt.). If I happen to be able to solve it again, I will quickly come to this thread and add the solution here so that I could always check back to my original line of thought, and I will keep my fingers crossed for that to occur!

Also, since I am new to the topic of Lagrange Multipliers, MarkFL offered to help guide me through using the technique on this problem.

First, he told me to state the objective function $f(x,y)$ and constraint function $g(x,y)=0$ and so I wrote:

\(\displaystyle f(x,y)=x+y\)

\(\displaystyle g(x,y)=x+y-3\left(\sqrt{x+1}+\sqrt{y+2} \right)=0\)

Next, he said to construct the following system:

\(\displaystyle f_x(x,y)=\lambda g_x(x,y)\)

\(\displaystyle f_y(x,y)=\lambda g_y(x,y)\)

And so I found:

\(\displaystyle 1=\lambda\left(1-\frac{3}{1\sqrt{x+1}} \right)\)

\(\displaystyle 1=\lambda\left(1-\frac{3}{1\sqrt{y+2}} \right)\)

And so I was then asked what this implies, and I could see that this implies:

\(\displaystyle x+1=y+2\)

\(\displaystyle y=x-1\)

So, Mark told me to substitute for $y$ in the constraint, and solve for $x$:

\(\displaystyle x+x-1-3\left(\sqrt{x+1}+\sqrt{x+1} \right)=0\)

\(\displaystyle 2x-1=6\sqrt{x+1}\)

Squaring both sides (keeping in mind extraneous solutions may be introduced):

\(\displaystyle 4x^2-4x+1=36x+36\)

\(\displaystyle 4x^2-40x-35=0\)

The quadratic formula gives us:

\(\displaystyle x=\frac{10\pm3\sqrt{15}}{2}\)

A check reveals that the smaller root is extraneous, so we are left with:

\(\displaystyle x=\frac{10+3\sqrt{15}}{2}\)

And so:

\(\displaystyle x+y=2x-1=9+3\sqrt{15}\)

And so we may conclude that one extremum is:

\(\displaystyle f\left(\frac{10+3\sqrt{15}}{2},\frac{8+3\sqrt{15}}{2} \right)=9+3\sqrt{15}\)

I asked Mark how we know this is a maximum, since we have found only one extremum. He said this is where the boundary extrema you suggested come into play. He told me to look at the constraint to see if it implies any further constraints on $x$ and $y$, and so I noticed we must have:

\(\displaystyle x+1\ge0\implies x\ge-1\)

\(\displaystyle y+2\ge0\implies y\ge-2\)

Letting $x=-1$, we find the constraint function implies:

\(\displaystyle -1+y-3\sqrt{y+2}=0\)

Solving for $y$, we then find:

\(\displaystyle y-1=3\sqrt{y+2}\)

\(\displaystyle y^2-2y+1=9y+18\)

\(\displaystyle y^2-11y-17=0\)

\(\displaystyle y=\frac{11\pm3\sqrt{21}}{2}\)

A check reveals that the smaller root is extraneous, so we are left with:

\(\displaystyle y=\frac{11+3\sqrt{21}}{2}\)

And so another critical point is:

\(\displaystyle \left(-1,\frac{11+3\sqrt{21}}{2} \right)\)

And we find:

\(\displaystyle f\left(-1,\frac{11+3\sqrt{21}}{2} \right)=\frac{3\left(3+\sqrt{21} \right)}{2}\)

Now, checking the other boundary, we may let $y=-2$ and the constraint implies:

\(\displaystyle x-2-3\sqrt{x+1}=0\)

\(\displaystyle x-2=3\sqrt{x+1}\)

\(\displaystyle x^2-4x+4=9x+9\)

\(\displaystyle x^2-13x-5=0\)

\(\displaystyle x=\frac{13\pm3\sqrt{21}}{2}\)

A check reveals that the smaller root is extraneous, so we are left with:

\(\displaystyle x=\frac{13+3\sqrt{21}}{2}\)

And so another critical point is:

\(\displaystyle \left(\frac{13+3\sqrt{21}}{2},-2 \right)\)

And we find:

\(\displaystyle f\left(\frac{13+3\sqrt{21}}{2},-2 \right)=\frac{3\left(3+\sqrt{21} \right)}{2}\)

Now, since:

\(\displaystyle \frac{3\left(3+\sqrt{21} \right)}{2}<9+3\sqrt{15}\)

we may then conclude:

\(\displaystyle f_{\min}=f\left(-1,\frac{11+3\sqrt{21}}{2} \right)=f\left(\frac{13+3\sqrt{21}}{2},-2 \right)=\frac{3\left(3+\sqrt{21} \right)}{2}\)

\(\displaystyle f_{\max}=f\left(\frac{10+3 \sqrt{15}}{2},\frac{8+3\sqrt{15}}{2} \right)=9+3\sqrt{15}\)

 

[Klaas van Aarsen]

Hello I like Serena,

Thanks for providing me such a great solution to tackle the problem using trigonometric approach and I appreciate it and I also like it!

Your patience and willingness to answer my "request" touched me very much. And I know I approached it differently than you (because I didn't introduce another variable ($\theta$ in your case) in my attempt.). If I happen to be able to solve it again, I will quickly come to this thread and add the solution here so that I could always check back to my original line of thought, and I will keep my fingers crossed for that to occur!

Also, since I am new to the topic of Lagrange Multipliers, MarkFL offered to help guide me through using the technique on this problem.

First, he told me to state the objective function $f(x,y)$ and constraint function $g(x,y)=0$ and so I wrote:

\(\displaystyle f(x,y)=x+y\)

\(\displaystyle g(x,y)=x+y-3\left(\sqrt{x+1}+\sqrt{y+2} \right)=0\)

Next, he said to construct the following system:

\(\displaystyle f_x(x,y)=\lambda g_x(x,y)\)

\(\displaystyle f_y(x,y)=\lambda g_y(x,y)\)

And so I found:

\(\displaystyle 1=\lambda\left(1-\frac{3}{1\sqrt{x+1}} \right)\)

\(\displaystyle 1=\lambda\left(1-\frac{3}{1\sqrt{y+2}} \right)\)

And so I was then asked what this implies, and I could see that this implies:

\(\displaystyle x+1=y+2\)

\(\displaystyle y=x-1\)

So, Mark told me to substitute for $y$ in the constraint, and solve for $x$:

\(\displaystyle x+x-1-3\left(\sqrt{x+1}+\sqrt{x+1} \right)=0\)

\(\displaystyle 2x-1=6\sqrt{x+1}\)

Squaring both sides (keeping in mind extraneous solutions may be introduced):

\(\displaystyle 4x^2-4x+1=36x+36\)

\(\displaystyle 4x^2-40x-35=0\)

The quadratic formula gives us:

\(\displaystyle x=\frac{10\pm3\sqrt{15}}{2}\)

A check reveals that the smaller root is extraneous, so we are left with:

\(\displaystyle x=\frac{10+3\sqrt{15}}{2}\)

And so:

\(\displaystyle x+y=2x-1=9+3\sqrt{15}\)

And so we may conclude that one extremum is:

\(\displaystyle f\left(\frac{10+3\sqrt{15}}{2},\frac{8+3\sqrt{15}}{2} \right)=9+3\sqrt{15}\)

I asked Mark how we know this is a maximum, since we have found only one extremum. He said this is where the boundary extrema you suggested come into play. He told me to look at the constraint to see if it implies any further constraints on $x$ and $y$, and so I noticed we must have:

\(\displaystyle x+1\ge0\implies x\ge-1\)

\(\displaystyle y+2\ge0\implies y\ge-2\)

Letting $x=-1$, we find the constraint function implies:

\(\displaystyle -1+y-3\sqrt{y+2}=0\)

Solving for $y$, we then find:

\(\displaystyle y-1=3\sqrt{y+2}\)

\(\displaystyle y^2-2y+1=9y+18\)

\(\displaystyle y^2-11y-17=0\)

\(\displaystyle y=\frac{11\pm3\sqrt{21}}{2}\)

A check reveals that the smaller root is extraneous, so we are left with:

\(\displaystyle y=\frac{11+3\sqrt{21}}{2}\)

And so another critical point is:

\(\displaystyle \left(-1,\frac{11+3\sqrt{21}}{2} \right)\)

And we find:

\(\displaystyle f\left(-1,\frac{11+3\sqrt{21}}{2} \right)=\frac{3\left(3+\sqrt{21} \right)}{2}\)

Now, checking the other boundary, we may let $y=-2$ and the constraint implies:

\(\displaystyle x-2-3\sqrt{x+1}=0\)

\(\displaystyle x-2=3\sqrt{x+1}\)

\(\displaystyle x^2-4x+4=9x+9\)

\(\displaystyle x^2-13x-5=0\)

\(\displaystyle x=\frac{13\pm3\sqrt{21}}{2}\)

A check reveals that the smaller root is extraneous, so we are left with:

\(\displaystyle x=\frac{13+3\sqrt{21}}{2}\)

And so another critical point is:

\(\displaystyle \left(\frac{13+3\sqrt{21}}{2},-2 \right)\)

And we find:

\(\displaystyle f\left(\frac{13+3\sqrt{21}}{2},-2 \right)=\frac{3\left(3+\sqrt{21} \right)}{2}\)

Now, since:

\(\displaystyle \frac{3\left(3+\sqrt{21} \right)}{2}<9+3\sqrt{15}\)

we may then conclude:

\(\displaystyle f_{\min}=f\left(-1,\frac{11+3\sqrt{21}}{2} \right)=f\left(\frac{13+3\sqrt{21}}{2},-2 \right)=\frac{3\left(3+\sqrt{21} \right)}{2}\)

\(\displaystyle f_{\max}=f\left(\frac{10+3 \sqrt{15}}{2},\frac{8+3\sqrt{15}}{2} \right)=9+3\sqrt{15}\)

Looking good!
Welcome to the Lagrange Multipliers Group.

 

[anemone]

Looking good!
Welcome to the Lagrange Multipliers Group.

I suspect the words "Lagrange Multipliers Group" that are highlighted in pink have some significant meaning, but I am unable to decipher the pink code, hehehe...

 

[Klaas van Aarsen]

I suspect the words "Lagrange Multipliers Group" that are highlighted in pink have some significant meaning, but I am unable to decipher the pink code, hehehe...

It does.

You can find more about the group and the secret of the pink code here, courtesy of Fernando Revilla.