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

In summary, the task involves determining the optimal values of variables a and b to either maximize or minimize a specific mathematical expression. This typically requires applying techniques such as calculus, including finding derivatives and solving equations, to identify critical points, followed by evaluating the expression to ascertain whether these points correspond to maxima or minima.
  • #1
songoku
2,384
351
Homework Statement
Please see below
Relevant Equations
not sure
1715840961999.png


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
 
Physics news on Phys.org
  • #2
The question is, which ##a## and ##b## minimize ##max(3a^2+2b,3b^2+2a)##.
 
  • Like
Likes songoku, MatinSAR, WWGD and 3 others
  • #3
Hill said:
The question is, which ##a## and ##b## minimize ##max(3a^2+2b,3b^2+2a)##.
Maybe related to Game Theory, Minimax Theorem?
 
  • Like
Likes songoku
  • #4
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.
 
  • Like
Likes songoku, SammyS, FactChecker and 1 other person
  • #5
pasmith said:
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.
 
  • Like
Likes songoku, FactChecker and PeroK
  • #6
Somehow the respective ( total)derivatives are "symmetrical": ## 6a +2 ; 6b+2##, though not sure what that implies here.
 
  • #7
The thread is in "precalculus." Are derivatives allowed?
 
  • #8
Hill said:
The thread is in "precalculus." Are derivatives allowed?
Good point. Not sure.
 
  • #9
Trial and error is definitely precalculus.
 
  • Like
Likes songoku and SammyS
  • #10
Hill said:
The thread is in "precalculus." Are derivatives allowed?

WWGD said:
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.
 
  • Like
Likes WWGD
  • #11
pasmith said:
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##.
 
  • Like
Likes songoku
  • #12
docnet said:
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.
 
  • Like
Likes songoku
  • #13
Hill said:
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

pasmith said:
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]?

pasmith said:
and only one of the given options satisfies this.
I think option (C) and (D) are still possible?

Thanks
 
  • #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##.
 
  • Like
Likes songoku
  • #15
songoku said:
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 .
 
  • Like
Likes songoku
  • #16
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)##.
 
  • #17
Hill said:
If one is allowed to go by the given options, then one only needs to compare the numbers calculated in the OP.
Exactly this.

songoku said:
I think option (C) and (D) are still possible?
What is the minimum of ## \left\{ \frac 1 3, \frac 8 {27} \right\} ##?

pasmith said:
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

pasmith said:
- and only one of the given options satisfies this.
is not correct: both (C) and (D) satisfy ## a = b ##.
 
  • Like
Likes docnet
  • #18
pbuk said:
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
 
  • Like
Likes PeroK
  • #19
docnet said:
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.
 
  • #20
PeroK said:
I wasn't sure how watertight the other proposed solutions were, hence post #14.
PeroK said:
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. :oops: I have to go to my job now.
 
  • #21
docnet said:
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##.
 
  • Like
Likes SammyS and docnet
  • #22
PeroK said:
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.
 

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

What is the general approach to find the maximum or minimum values of an expression involving variables a and b?

The general approach involves taking the derivative of the expression with respect to the variables a and b. You then set the derivatives equal to zero to find critical points. After identifying these points, you can use the second derivative test or evaluate the expression at these points to determine whether they correspond to a maximum, minimum, or saddle point.

How do I determine if the critical points correspond to a maximum or minimum?

You can determine the nature of the critical points by using the second derivative test. For a function of two variables, you calculate the Hessian matrix at the critical points. If the determinant of the Hessian is positive and the second derivative with respect to a (or b) is positive, then you have a local minimum. If the determinant is positive and the second derivative is negative, then you have a local maximum. If the determinant is negative, the point is a saddle point.

What if the expression is constrained by certain conditions?

If the expression is constrained, you can use the method of Lagrange multipliers. This involves introducing a new variable (the Lagrange multiplier) and setting up a system of equations that includes the original function and the constraint. By solving this system, you can find the values of a and b that maximize or minimize the expression under the given constraints.

Can I find maximum or minimum values using graphical methods?

Yes, graphical methods can be useful, especially for visualizing the behavior of the expression. By plotting the function in 2D or 3D, you can identify peaks and valleys that correspond to maximum and minimum values. However, this method may not provide precise values and is generally used as a supplementary approach to analytical methods.

What role does the domain of the variables play in finding max/min values?

The domain of the variables a and b is crucial as it defines the feasible region where you can search for maximum or minimum values. If the domain is restricted (e.g., a and b must be non-negative), it can affect the location of critical points and the overall behavior of the expression. Always ensure that the critical points found are within the specified domain before concluding about their maximum or minimum status.

Similar threads

Back
Top