Find the values of a and b to get max/min of certain expression

[songoku]

Homework Statement

Please see below

Relevant Equations

not sure

I don't think I understand what the question is asking. Find ##a## and ##b## so that the maximum takes the minimum values?

If ##a## and ##b## are real numbers, the maximum values of ##3a^2+2b## and ##3b^2+2a## are infinity. How can infinity take minimum value?

I have calculated the values of ##3a^2+2b## and ##3b^2+2a## for each given option:
Option (A): ##3a^2+2b=-\frac{1}{9}## and ##3b^2+2a=-\frac{14}{27}##
Option (B): ##3a^2+2b=-\frac{14}{27}## and ##3b^2+2a=-\frac{1}{9}##
Option (C): ##3a^2+2b=-\frac{8}{27}## and ##3b^2+2a=-\frac{8}{27}##
Option (D): ##3a^2+2b=-\frac{1}{3}## and ##3b^2+2a=-\frac{1}{3}##
Option (E): ##3a^2+2b=-\frac{1}{3}## and ##3b^2+2a=1##

What do this values tell me and how should I actually interpret the question?

Thanks

 

[Hill]

The question is, which ##a## and ##b## minimize ##max(3a^2+2b,3b^2+2a)##.

 

[WWGD]

The question is, which ##a## and ##b## minimize ##max(3a^2+2b,3b^2+2a)##.

Maybe related to Game Theory, Minimax Theorem?

 

[pasmith]

Note that sapping [itex]a[/itex] and [itex]b[/itex] swaps [itex]3a^2 + 2b[/itex] and [itex]3b^2 + 2a[/itex], leaving the maximum invariant. Note also that [itex]3a^2 + 2b[/itex] is strictly increasing in the direction of increasing [itex]b[/itex], and [itex]3b^2 + 2a[/itex] is strictly increasing in the direction of increasing [itex]a[/itex].

These observations suggest that [itex]\min_{(a,b)} \max \{3a^2 + 2b, 3b^2 + 2a\}[/itex] lies on the line [itex]a = b[/itex] - and only one of the given options satisfies this.

 

[Hill]

Note that sapping [itex]a[/itex] and [itex]b[/itex] swaps [itex]3a^2 + 2b[/itex] and [itex]3b^2 + 2a[/itex], leaving the maximum invariant. Note also that [itex]3a^2 + 2b[/itex] is strictly increasing in the direction of increasing [itex]b[/itex], and [itex]3b^2 + 2a[/itex] is strictly increasing in the direction of increasing [itex]a[/itex].

These observations suggest that [itex]\min_{(a,b)} \max \{3a^2 + 2b, 3b^2 + 2a\}[/itex] lies on the line [itex]a = b[/itex] - and only one of the given options satisfies this.

If one is allowed to go by the given options, then one only needs to compare the numbers calculated in the OP.

 

[WWGD]

Somehow the respective ( total)derivatives are "symmetrical": ## 6a +2 ; 6b+2##, though not sure what that implies here.

 

[Hill]

The thread is in "precalculus." Are derivatives allowed?

 

[WWGD]

The thread is in "precalculus." Are derivatives allowed?

Good point. Not sure.

 

[PeroK]

Trial and error is definitely precalculus.

 

[Mark44]

The thread is in "precalculus." Are derivatives allowed?

Good point. Not sure.

Unless a question posted in this section is clearly calculus-related (and thereby misplaced), replies should not bring derivatives or integrals to bear.

 

[docnet]

Note that sapping [itex]a[/itex] and [itex]b[/itex] swaps [itex]3a^2 + 2b[/itex] and [itex]3b^2 + 2a[/itex], leaving the maximum invariant. Note also that [itex]3a^2 + 2b[/itex] is strictly increasing in the direction of increasing [itex]b[/itex], and [itex]3b^2 + 2a[/itex] is strictly increasing in the direction of increasing [itex]a[/itex].

These observations suggest that [itex]\min_{(a,b)} \max \{3a^2 + 2b, 3b^2 + 2a\}[/itex] lies on the line [itex]a = b[/itex] - and only one of the given options satisfies this.

Awesome! To add, if ##a=b##, then
$$3a^2+2b=3b^2+2a=3a^2+2a.$$
And the global minimum of ##3a^2 + 2a## for all ##a\in \mathbb{R}## is $$3\cdot \frac{1}{(-3)^2}+3\cdot \frac{1}{-3}=-\frac{1}{3}.$$ This indeed seems to be the minimum value of [itex]\max \{3a^2 + 2b, 3b^2 + 2a\}[/itex] considering all ##a,b\in \mathbb{R}##.

However, the implied conclusion is not perfectly right because the condition '##a=b##' doesn't guarantee the minimum quantity of [itex]\max \{3a^2 + 2b, 3b^2 + 2a\}[/itex]. It depends on the given choices of ##a## and ##b##.

 

[Hill]

Awesome! To add, if ##a=b##, then
$$3a^2+2b=3b^2+2a=3a^2+2a.$$
And the global minimum of ##3a^2 + 2a## for all ##a\in \mathbb{R}## is $$3\cdot \frac{1}{(-3)^2}+3\cdot \frac{1}{-3}=-\frac{1}{3}.$$ This indeed seems to be the minimum value of [itex]\max \{3a^2 + 2b, 3b^2 + 2a\}[/itex] considering all ##a,b\in \mathbb{R}##.

However, the implied conclusion is not perfectly right because the condition '##a=b##' doesn't guarantee the minimum quantity of [itex]\max \{3a^2 + 2b, 3b^2 + 2a\}[/itex]. It depends on the given choices of ##a## and ##b##.

To minimize ##\max \{3a^2 + 2b, 3b^2 + 2a\}## is the same as to minimize ##\max \{(3a^2 + 2b)-(3b^2 + 2a), 0\}##.
The latter is minimized when ##0=(3a^2 + 2b)-(3b^2 + 2a)=3(a^2-b^2) - 2(a-b)=(a-b)(3(a+b)-2)##.
This happens when either ##a-b=0## or ##3(a+b)-2=0##.
So, the minimum happens when either ##3a^2 + 2a## or ##3a^2 + 2(2/3-a)## is minimized.
The mins of the last two expression are easy to find and then we take the smaller of the two.

 

[songoku]

To minimize ##\max \{3a^2 + 2b, 3b^2 + 2a\}## is the same as to minimize ##\max \{(3a^2 + 2b)-(3b^2 + 2a), 0\}##.
The latter is minimized when ##0=(3a^2 + 2b)-(3b^2 + 2a)=3(a^2-b^2) - 2(a-b)=(a-b)(3(a+b)-2)##.
This happens when either ##a-b=0## or ##3(a+b)-2=0##.
So, the minimum happens when either ##3a^2 + 2a## or ##3a^2 + 2(2/3-a)## is minimized.
The mins of the last two expression are easy to find and then we take the smaller of the two.

I understand

Note that sapping [itex]a[/itex] and [itex]b[/itex] swaps [itex]3a^2 + 2b[/itex] and [itex]3b^2 + 2a[/itex], leaving the maximum invariant. Note also that [itex]3a^2 + 2b[/itex] is strictly increasing in the direction of increasing [itex]b[/itex], and [itex]3b^2 + 2a[/itex] is strictly increasing in the direction of increasing [itex]a[/itex].

These observations suggest that [itex]\min_{(a,b)} \max \{3a^2 + 2b, 3b^2 + 2a\}[/itex] lies on the line [itex]a = b[/itex]

I understand the 1st paragraph but sorry I don't understand the 2nd one. How can we know the answer will lie on the line [itex]a = b[/itex]? How can we draw this conclusion from the fact [itex]3a^2 + 2b[/itex] is strictly increasing in the direction of increasing [itex]b[/itex], and [itex]3b^2 + 2a[/itex] is strictly increasing in the direction of increasing [itex]a[/itex]?

and only one of the given options satisfies this.

I think option (C) and (D) are still possible?

Thanks

 

[PeroK]

Here's an alternative approach. Note that we must have ##a, b < 0##. As both expressions are less in this case than for the equivalent positive numbers. We can take ##c = -a, d = -b##, with ##c, d > 0## and minimise the maximum of: ##3c^2 - 2d## and ##3d^2 - 2c##. Now, for ##c, d > 0## we have:
$$3c^2 -2d \ge 3d^2 - 2c \ \iff \ c \ge d$$You can check that out. Using the symmetry, that lets us minimise ##3c^2 - 2d## for ##c \ge d > 0##. This is clearly minimised when ##c = d##. Finally, therefore, we minimise ##3c^2 - 2c## for ##c > 0##. Then, the solution to the original problem is ##a = b = -c##.

 

[SammyS]

I understand the 1st paragraph but sorry I don't understand the 2nd one. How can we know the answer will lie on the line [itex]a = b[/itex]? How can we draw this conclusion from the fact [itex]3a^2 + 2b[/itex] is strictly increasing in the direction of increasing [itex]b[/itex], and [itex]3b^2 + 2a[/itex] is strictly increasing in the direction of increasing [itex]a[/itex]?

I think option (C) and (D) are still possible?

Thanks

You wrote the above in response to @pasmith's Post #4 .

Maybe adding a few clarifying words will help you understand.

Keeping ##a## fixed, ##3a^2 + 2b## is strictly increasing in the direction of increasing ##b##, and keeping ##b## fixed, ##3b^2 + 2a## is strictly increasing in the direction of increasing ##a##. Therefore, one of ##3a^2 + 2b## or ##3b^2 + 2a## is greater than the lesser of ##3a^2 + 2a## or ##3b^2 + 2b##.

It follows that if there is some number, ##R##, such that ##3r^2+2r## is a minimum at ##r=R##, then it seems you have a solution. ##a=b=R## .

You shouldn't need to use Calculus if you understand quadratic functions .

 

[docnet]

To be honest, what I'm struggling to understand is that if one's given several choices for ##(a,b)##, including one choice such that ##a=b##, the value of ## \max \{3a^2 + 2b, 3b^2 + 2a\}## with ##a=b## is not necessarily be the minimum over all possible choices of ##(a,b)##.

 

[pbuk]

If one is allowed to go by the given options, then one only needs to compare the numbers calculated in the OP.

Exactly this.

I think option (C) and (D) are still possible?

What is the minimum of ## \left\{ \frac 1 3, \frac 8 {27} \right\} ##?

These observations suggest that [itex]\min_{(a,b)} \max \{3a^2 + 2b, 3b^2 + 2a\}[/itex] lies on the line [itex]a = b[/itex]

This is true, but it is not necessary to observe this to answer the question, and nor is it sufficient because

- and only one of the given options satisfies this.

is not correct: both (C) and (D) satisfy ## a = b ##.

 

[docnet]

This is true, but it is not necessary to observe this to answer the question, and nor is it sufficient

Which is what I've been trying to say as well, without meaning to sound snarky. hehe

 

[PeroK]

Which is what I've been trying to say as well, without meaning to sound snarky. hehe

I wasn't sure how watertight the other proposed solutions were, hence post #14.

 

[docnet]

I wasn't sure how watertight the other proposed solutions were, hence post #14.

Here's an alternative approach. Note that we must have ##a, b < 0##. As both expressions are less in this case than for the equivalent positive numbers. We can take ##c = -a, d = -b##, with ##c, d > 0## and minimise the maximum of: ##3c^2 - 2d## and ##3d^2 - 2c##. Now, for ##c, d > 0## we have:
$$3c^2 -2d \ge 3d^2 - 2c \ \iff \ c \ge d$$You can check that out. Using the symmetry, that lets us minimise ##3c^2 - 2d## for ##c \ge d > 0##. This is clearly minimised when ##c = d##. Finally, therefore, we minimise ##3c^2 - 2c## for ##c > 0##. Then, the solution to the original problem is ##a = b = -c##.

I'm, sorry it quite right either. And I humbly suggest why: you found a condition on ##a## and ##b## that minimizes the expression ##\min_{(a,b)} \max \{3a^2 + 2b, 3b^2 + 2a\}##, by assuming one could freely choose ##a## and ##b## from all of ##\mathbb{R}^2##. It's much simpler problem to the general case where ##(a,b)## is restricted to an arbitrary subset of ##\mathbb{R}^2##. Note the condition "##a=b##" alone is insufficient for finding the desired pair ##a,b##. And the simplest example I could think of is $$\{(1,1),(-\frac{1}{2},\frac{1}{2})\},$$ where the first pair is not the optimal choice.

I don't know if it's possible to define an algorithm to find the right ##(a,b)## in the general case. I have a feeling that there may even exist subsets of ##\mathbb{R}^2## that makes the exact solution impossible to find, even if it could be proven to exist.

Edit: The problem in the OP was probably ok with a simple comparison-based answer, and it didn't even ask to show work. I have to go to my job now.

 

[PeroK]

I'm, sorry it quite right either. And I humbly suggest why: you found a condition on ##a## and ##b## that minimizes the expression ##\min_{(a,b)} \max \{3a^2 + 2b, 3b^2 + 2a\}##, by assuming one could freely choose ##a## and ##b## from all of ##\mathbb{R}^2##. It's much simpler problem to the general case where ##(a,b)## is restricted to an arbitrary subset of ##\mathbb{R}^2##. Note the condition "##a=b##" alone is insufficient for finding the desired pair ##a,b##. And the simplest example I could think of is $$\{(1,1),(-\frac{1}{2},\frac{1}{2})\},$$ where the first pair is not the optimal choice.

I don't know if it's possible to define an algorithm to find the right ##(a,b)## in the general case. I have a feeling that there may even exist subsets of ##\mathbb{R}^2## that makes the exact solution impossible to find, even if it could be proven to exist.

The solution I presented was to minimise the maximum across all real values of ##a, b##, as implied in the original question. This question itself was short-circuited by being a multiple choice, where simple trial and error is sufficient.

I wasn't looking at more general cases, where ##a, b## are constrained to a subset of ##\mathbb R^2##.

 

[docnet]

The solution I presented was to minimise the maximum across all real values of ##a, b##, as implied in the original question. This question itself was short-circuited by being a multiple choice, where simple trial and error is sufficient.

I wasn't looking at more general cases, where ##a, b## are constrained to a subset of ##\mathbb R^2##.

doh! I somehow ignored the original problem statement and instead considered problem I thought was more interesting. I'm an aloof and certified dork. Oof.