Problem Solving

Hi Rahul and fabio,
Does the differentiation method always lead to a correct answer ?
Can you confirm ? Can you also give a good example when this can
be used and this should not be used ?


Thanks,
TOPGMAT

TOPGMAT wrote:Hi Rahul and fabio,
Does the differentiation method always lead to a correct answer ?
Can you confirm ? Can you also give a good example when this can
be used and this should not be used ?


Thanks,
TOPGMAT

Hi, TOPGMAT!

As a "rule", you may use "differentiate and equal to zero" to find extremal points of (differentiable) functions of a single variable, that is, f = f(x) but, as I said, do not try to adapt that (in a "naive way") to f = f(x,y) functions, for instance (when y is independent of x, for sure).

Apart from that, everytime people do that ("differentiate and equal to zero") in the GMAT-context properly (in the context mentioned above) they could argument as Rahul did (in terms of even powers are always non-negative) or (very common) exploring some 2o degree single-variable expression property, mainly through the vertex of a parabola (point at which we have minimum or maximum values, according to positive or negative concavity)...

Examples of applications: have a look at my BTG posts contributions:

01. Link: https://www.beatthegmat.com/value-of-s-w ... tml#314200

02. Link: https://www.beatthegmat.com/is-x-2-x-1-m ... tml#309911

for instance.

I hope you got the point and, of course, that I answered your question satisfactory.

Regards,
Fabio.

Hi Fabio,
In a f(x,y) equation....
Can we first differentiate wrt to x and then wrt y and add up ????
I wanted to know whether this holds good for all cases...



Also, if f(x,y) = xy^2 + yx^2.
Can we proceed with the above method ?

Thanks,
TOPGMAT

Hi, TOPGMAT!

I guess here is not the proper place to deal with Calculus of functions of two-real variables but... let´s talk a bit about it, no problem. (Thank you for the "thanks", by the way.)

Let us consider that f=f(x,y) is defined for all the plane RxR, as it is the case with the explicit f=f(x,y) function you chose. (Good one for the discussions, let´s focus on it!!)

When we consider (say) y fixed [x fixed] and we differentiate f related to the x variable [y variable] , we are calculating the partial x-derivative [y-derivative] of f. Let us called them D1f (for x) and D2f (for y), ok?

In your example, D1f(x,y) = y^2 + 2xy and D2f(x,y) = x^2 + 2xy. (Hope you remember this calculations.)

From the fact that D1f and D2f are continuous functions in every (x,y) point of RxR, there is a theorem that assures us that f = f(x,y) is differentiable everywhere, that is, in the whole RxR plane. From the fact that the partial derivatives of D1f and D2f (viewed simply as functions of x and y themselves) are also continuous in the plane, and so on, we are sure the function you chose is (called C^infinity and)is really great in terms of "good behavior" and in terms of being able to calculate D1(D1f) we will call D11f, D2(D2f) we will call D22f and the "mixed" ones, D1(D2f) and D2(D1f), not to mention the next ones D111f, etc (and infinitum)...

Besides that, whenever we have D1f(a,b) = D2f(a,b) = 0, we say (a,b) is a critical point of f and it is known that a NECESSARY condition for f to have a local extremum at (a,b) is that (a,b) is a critical point of f.

The problem is that this is not SUFFICIENT, in other words, we have to consider the second derivatives (I mean D11f, D12f, etc) to determine whether we have a local maximum, a local minimum or yet a (so called) saddle point.As the name suggests, in a neighborhood of a saddle point you have points where the function gets smaller and greater values than in itself and therefore, no minimum or maximum is attained in saddle points.

A typical test to decide that (minimum/maximum/saddle) is the so-called "second-derivative" test and this test is related to the "hessian matrix" (google about it, if you want additional info). This is the matrix I had in mind in my previous post, by the way...

In the function you chose, from the fact that D11f(0,0)*D22f(0,0) - D12f(0,0) = 0, the test gives no conclusion (we say the test is "inconclusive"), but it is not hard to guess that (0,0) is a saddle point, because if you take (say) (x,y)=(0.001, 0.001) it is easy to see that f(x,y) is greater than 0 (therefore, in a non-rigourous way, you see that the origin is not a local maximum because you could use 0.0001 and 0.0001... got it?) and if you take (x,y) = (-0.001, -0.001) then f(x,y) is less than 0, and now in a very non-rigorous way you see that the origin is also not a local minimum (same idea)... therefore the origin is a saddle point.

Your example is, therefore, one good counter-example to people who believe that "(partial) differentiate and equal to zero" will get (always) a maximum or minimum point for the function in question, but it is true that, when this operations are viable (when f is at least twice differentiable it´s the case), the critical points you may obtain are the ONLY "candidates" for maximum/minimum, for sure! (So you could "test" one by one, by some methods as the hessian analysis, or even as stupid as mine, trying to calculate values "near" the critical point considered.)

In summary: the single (real) variable scenario is much easily "tamed" than the more-than-one variable scenario, therefore I guess my suggestion should be taken seriously: avoid using Calculus on the GMAT because it is certainly not necessarily and, sometimes, may lead you to some subtle details that, in general, the candidate will not remember nor have enough time to deal with during the question, not to mention some wrong conclusions he/she might take.

I hope you got the point and please apology any lack of rigorous on my part: now is saturday night here in Brazil, and my wife is asking me to see a DVD with her for a couple of minutes already...

Regards,
Fabio.

Thanks fabio.
I hope you enjoyed the movie...
BTW your explanation was as good as a GMAT RC

Conclusion
1) f(x) = x^2+ .... (single variable function)... We can use differentiation.
Take second derivative and check x value results is a value less than 0.

2) f(x,y) functions... don't use calculus atleast for gmat .
There will be surely an easy alternative.

Thanks for the guidance.

TOPGMAT wrote:Thanks fabio.
I hope you enjoyed the movie...
BTW your explanation was as good as a GMAT RC

Conclusion
1) f(x) = x^2+ .... (single variable function)... We can use differentiation.
Take second derivative and check x value results is a value less than 0.

2) f(x,y) functions... don't use calculus atleast for gmat .
There will be surely an easy alternative.

Thanks for the guidance.

Thank you for the nice compliments, TOPGMAT!
Regards,
Fabio.

P.S.: yep, VERY good film (with Darin, the great Argentinian actor)... I guess the name of the film was "Lo segredo de tu ojos" or something like that.

Rahul@gurome wrote:

thp510 wrote:...
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18

To use the differentiation rule, you need to know differential calculus which is not a part of GMAT. You can proceed as I mentioned earlier. Let's do it again. To minimize the expression, we have to minimize each term of it. Simple algebraical method to do it is to rearrange the terms in such a way that we get square terms. This is because minimum value of a square term is zero.

1. # 2x² + 3y² - 4x - 12y + 18
2. = 2(x² - 2x) + 3(y² - 4y) + 18 .................... What you've done
3. = 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
4. = 2(x - 1)² + 3(y - 2)² + 4

For minimum value of the expression, 2(x - 1)² and 3(y - 2)² must be minimum, i.e. equal to zero. Thus minimum value of the expression is 4.

I have 2 stupid questions to ask.

FIRST
In step 3, we broke down "18" as 2+12+4. How did we break this down to the numbers mentioned?

We could have also done 4+6+8, which would have changed the expression to 8. Is it not ?



SECOND
How do we find factors...like (x^2 - 2x + 1).....as (x - 1)^2 ?


Thanks!

Rahul@gurome wrote:

thp510 wrote:...
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18

To use the differentiation rule, you need to know differential calculus which is not a part of GMAT. You can proceed as I mentioned earlier. Let's do it again. To minimize the expression, we have to minimize each term of it. Simple algebraical method to do it is to rearrange the terms in such a way that we get square terms. This is because minimum value of a square term is zero.

# 2x² + 3y² - 4x - 12y + 18
= 2(x² - 2x) + 3(y² - 4y) + 18 .................... What you've done
= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4

For minimum value of the expression, 2(x - 1)² and 3(y - 2)² must be minimum, i.e. equal to zero. Thus minimum value of the expression is 4.

I am sorry I still dont understand why 18 becomes 4 here. I understand the concept of pulling together like terms and then factoring out common items. But when I do it to me it feel like it should go:

2x(x-2) + 3y(y-4) + 18 --- still not sure how the min value here is 4. Would you be willing to explain again?

winnerhere wrote:Find the minimum value of an expression

2 (x^2) + 3 (y^2) - 4x - 12y + 18

1) 18

2)10

3) 4

4) 0

5) -10

I received a PM asking me to comment. Let's start by reorganizing the expression.

2 (x^2) + 3 (y^2) - 4x - 12y + 18

= 2x^2 - 4x + 3y^2 - 12y + 18

= 2(x^2 - 2x) + 3(y^2 - 4y) + 18

Let's handle each part of the expression separately, starting with (x^2 - 2x).
We need to complete the square inside the (). We're looking for a value that when added to itself will yield a sum of -2 (the coefficient of -2x).
Thus, the needed value inside the parentheses is -1:

(x-1)^2 = x^2 - 2x + 1

Now in the expression 2(x^2 - 2x) + 3(y^2 - 4y) + 18 we replace (x^2 - 2x) with (x^2 - 2x +1):

2(x^2 - 2x + 1) + 3(y^2 - 4y) + 18

But we can't change the value of the expression. Since we placed +1 inside the (), and the entire expression inside the () is being multiplied by 2, we need to subtract 2*1 = 2 from the whole expression so that its value is unchanged:

2(x^2 - 2x + 1) + 3(y^2 - 4y) + 18 - 2

= 2(x-1)^2 + 3(y^2 - 4y) + 16

Now let's handle the second part of the expression, (y^2 - 4y).

We need to complete the square inside the (). We're looking for a value that when added to itself will yield a sum of -4 (the coefficient of -4y). Thus, the needed value inside the parentheses is -2:

(y-2)^2 = y^2 - 4y + 4

Now in the expression 2(x-1)^2 + 3(y^2 - 4y) + 16 we replace (y^2 - 4y) with (y^2 - 4y + 4):

2(x-1)^2 + 3(y^2 - 4y + 4) + 16

But we can't change the value of the expression. Since we placed +4 inside the (), and the entire expression inside the () is being multiplied by 3, we need to subtract 3*4 = 12 from the whole expression so that the value of the expression is unchanged:

2(x-1)^2 + 3(y^2 - 4y + 4) + 16 - 12

= 2(x-1)^2 + 3(y-2)^2 + 4.

Now we can determine the minimum value of the expression. The expression will be minimized when x=1 and y=2 so that the first two terms equal 0:

2(x-1)^2 + 3(y-2)^2 + 4 = 2(1-1)^2 + 3(2-2)^2 + 4 = 0 + 0 + 4 = 4.

The correct answer is C.

Hope this helps!

Hi , I have done this by the below way
2x^2+3y^2-4x-12y+18
Let consider

2x^2+3y^2-4x-12y+18=0
2x^2-4x+2+3y^2-12y+12+4=0
2(x-1)^2+3(y-2)^+4=0

So X=1 & Y=2

Then value is 4

the Ans C

I did it quite quick so I won't need to type a page long explanation.

Step 1. Break into 3 pieces:

1. 2x^2 - 4x

2. 3y^2 - 12y

3. 18

Step 2. Take a closer look at the first two

Signs. We know that negative variables won't work, because square of a negative is positive, and minus a negative is positive. -> answer if not 0 is a positive number

Size. The higher the number, we plug in, it is obvious that the first half of each equation will be much bigger, since we are squaring. -> a small positive number is what we need

Step 3. Plug in small positive numbers

Plugging in 1 and 2 for first one is enough for a conclusion, that 1 is what we are looking for with the smallest result of -2.
Plugging in 1 and 2 and 3 for the second one, we realize quickly 2 is what we are looking for with the smallest result of -12.

Step 4.

(-2)+(-12)+18=4

winnerhere wrote:Find the minimum value of an expression

2 (x^2) + 3 (y^2) - 4x - 12y + 18

1) 18

2)10

3) 4

4) 0

5) -10

2 (x^2) + 3 (y^2) - 4x - 12y + 18

= (x √2) ^2 - 2 (x √2) (√2) + (√2) ^2 + (y √3) ^2 - 2 (y √3) (2 √3) + (2 √3) ^2 + 4

= (x √2 - √2) ^2 + (y √3 - 2 √3) ^2 + 4

It is minimum at x = 1 and y = 2, and the minimum value is [spoiler]4.



Choice (3)[/spoiler]

Zerks87 wrote:

Rahul@gurome wrote:

thp510 wrote:...
Not getting it. What does the equation look like before you get to 4x=4 or 6y=12 using the differentiation rule. I'm still stuck at the following:

2(x^2-2x)+3(y^2-4y)+18

To use the differentiation rule, you need to know differential calculus which is not a part of GMAT. You can proceed as I mentioned earlier. Let's do it again. To minimize the expression, we have to minimize each term of it. Simple algebraical method to do it is to rearrange the terms in such a way that we get square terms. This is because minimum value of a square term is zero.

# 2x² + 3y² - 4x - 12y + 18
= 2(x² - 2x) + 3(y² - 4y) + 18 .................... What you've done
= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4

For minimum value of the expression, 2(x - 1)² and 3(y - 2)² must be minimum, i.e. equal to zero. Thus minimum value of the expression is 4.

I am sorry I still dont understand why 18 becomes 4 here. I understand the concept of pulling together like terms and then factoring out common items. But when I do it to me it feel like it should go:

2x(x-2) + 3y(y-4) + 18 --- still not sure how the min value here is 4. Would you be willing to explain again?

We have +18 as a constant given, (√2) ^2 takes 2 out of it and leaves +16 and then (2 √3) ^2 takes 12 out of it and leaves

JUST 4u

Thanks. Your explanation really made the solution simple.

GMATGuruNY wrote:

winnerhere wrote:Find the minimum value of an expression

2 (x^2) + 3 (y^2) - 4x - 12y + 18

1) 18

2)10

3) 4

4) 0

5) -10

I received a PM asking me to comment. Let's start by reorganizing the expression.

2 (x^2) + 3 (y^2) - 4x - 12y + 18

= 2x^2 - 4x + 3y^2 - 12y + 18

= 2(x^2 - 2x) + 3(y^2 - 4y) + 18

Let's handle each part of the expression separately, starting with (x^2 - 2x).
We need to complete the square inside the (). We're looking for a value that when added to itself will yield a sum of -2 (the coefficient of -2x).
Thus, the needed value inside the parentheses is -1:

(x-1)^2 = x^2 - 2x + 1

Now in the expression 2(x^2 - 2x) + 3(y^2 - 4y) + 18 we replace (x^2 - 2x) with (x^2 - 2x +1):

2(x^2 - 2x + 1) + 3(y^2 - 4y) + 18

But we can't change the value of the expression. Since we placed +1 inside the (), and the entire expression inside the () is being multiplied by 2, we need to subtract 2*1 = 2 from the whole expression so that its value is unchanged:

2(x^2 - 2x + 1) + 3(y^2 - 4y) + 18 - 2

= 2(x-1)^2 + 3(y^2 - 4y) + 16

Now let's handle the second part of the expression, (y^2 - 4y).

We need to complete the square inside the (). We're looking for a value that when added to itself will yield a sum of -4 (the coefficient of -4y). Thus, the needed value inside the parentheses is -2:

(y-2)^2 = y^2 - 4y + 4

Now in the expression 2(x-1)^2 + 3(y^2 - 4y) + 16 we replace (y^2 - 4y) with (y^2 - 4y + 4):

2(x-1)^2 + 3(y^2 - 4y + 4) + 16

But we can't change the value of the expression. Since we placed +4 inside the (), and the entire expression inside the () is being multiplied by 3, we need to subtract 3*4 = 12 from the whole expression so that the value of the expression is unchanged:

2(x-1)^2 + 3(y^2 - 4y + 4) + 16 - 12

= 2(x-1)^2 + 3(y-2)^2 + 4.

Now we can determine the minimum value of the expression. The expression will be minimized when x=1 and y=2 so that the first two terms equal 0:

2(x-1)^2 + 3(y-2)^2 + 4 = 2(1-1)^2 + 3(2-2)^2 + 4 = 0 + 0 + 4 = 4.

The correct answer is C.

Hope this helps!

I tried to find out what the question was trying to find out
The question was testing several things it seemed to me (like do you know that 0 as an exponent yeilds 1 as an answer)
The MAIN thing the problem wanted to find out was whether I understood that raising a number to a positive power other than 1 or zero has more of a maginifying effect than multiplication. So most probabably x and y would have the same value.
To me it seemed the aim was to minimize the exponentials since that are more "powerful" than multiplication. I also saw that I had to simultaneoulsy maximize the multiplication in order to at least subtract something from 18.
I saw that if I used 0 for x and 0 for y I would not be doing any "damage" with the multiplication
I so I abandonded 0 as a choice immediately. I went on to 1 as a possible value for both x and y and came up with 4 as a possible answer.
I saw that using 2 would produce a bigger result than 4 because of the exponential vs multiplication effect I mentioned earlier SO 4 MUST be the answer. SO I chose C as the answer.
This solution took me 74 seconds.
PS I am minoring in Math but it would never have occured to me to use differentiation because it is beyond the scope of the GMAT. My motto is is the cliche KISS.
The main thing I have come to learn is to first think about what the question is trying to see if you know. Once you have a pretty good idea then the solution is often not that difficult. GMAT is more of an IQ test to me than a real Math test.