Minimum value of an expression
- Deependra1
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grouping x & y together, 2x^2 + 3y^2 -4x -12y +18 can be reorganized as
2x(x-2) + 3y(y-4) + 18
Inorder for the above to have a min value 2x(x-2) and 3y(y-4) should have max negative val as possible
For 2x(x-2)<0, x<2 but x cannot be negative. otherwise 2x(x-2) will become positive. So for x=1, this will have max negative value = -2
For 3y(y-4) < 0, y<4. Possible values for y is 3,2,1
For y=3, 3y(y-4) = -9
For y=2, 3y(y-4) = -12
for y=1, 3y(y-4) = -9
So min value is attained when y=2
So min val of the whole exp is when x=1 and y=2
Hence the min val is -2 -12 + 14 = -4
Ans is C
2x(x-2) + 3y(y-4) + 18
Inorder for the above to have a min value 2x(x-2) and 3y(y-4) should have max negative val as possible
For 2x(x-2)<0, x<2 but x cannot be negative. otherwise 2x(x-2) will become positive. So for x=1, this will have max negative value = -2
For 3y(y-4) < 0, y<4. Possible values for y is 3,2,1
For y=3, 3y(y-4) = -9
For y=2, 3y(y-4) = -12
for y=1, 3y(y-4) = -9
So min value is attained when y=2
So min val of the whole exp is when x=1 and y=2
Hence the min val is -2 -12 + 14 = -4
Ans is C
- jmckenna14
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= 2(x² - 2x + 1) + 3(y² - 4y + 4) + 4
= 2(x - 1)² + 3(y - 2)² + 4
I lost you....how did you get +1 and +4 in parentheses?
Can you walk all the way through the end of the answer?
= 2(x - 1)² + 3(y - 2)² + 4
I lost you....how did you get +1 and +4 in parentheses?
Can you walk all the way through the end of the answer?
Rahul@gurome wrote: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.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
# 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.
Hi Mitch,
Isnt it too complex to this way !! If we have to find the minimum value .. cant we pick x = 1, y = 2 and get the desired output. ?
Isnt it too complex to this way !! If we have to find the minimum value .. cant we pick x = 1, y = 2 and get the desired output. ?
GMATGuruNY wrote:I received a PM asking me to comment. Let's start by reorganizing the expression.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
= 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!
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Q: 2 (x^2) + 3 (y^2) - 4x - 12y + 18
There is no need for complexities such as Differential calculus or factorization. I remember reading somewhere quadratic equations with two variables need NOT be factorized on the GMAT to solve problems.
So, here is the SPLIT FUNCTION APPROACH.
Treat them as separate functions one of x and one of y
2(x^2) -4x
Try x = 0 --> 0
Try x = 1 --> -2 (min)
Try x = 2 -->0
3(y^2) -12y
Try y = 0 -->0
Try y = -1--> 15
Try y = 1 -->-9
Try y = 2 -->-12 (min)
Try y = 3 -->-9
Thus answer is (1,2) --> which yields answer as 4.
There is no need for complexities such as Differential calculus or factorization. I remember reading somewhere quadratic equations with two variables need NOT be factorized on the GMAT to solve problems.
So, here is the SPLIT FUNCTION APPROACH.
Treat them as separate functions one of x and one of y
2(x^2) -4x
Try x = 0 --> 0
Try x = 1 --> -2 (min)
Try x = 2 -->0
3(y^2) -12y
Try y = 0 -->0
Try y = -1--> 15
Try y = 1 -->-9
Try y = 2 -->-12 (min)
Try y = 3 -->-9
Thus answer is (1,2) --> which yields answer as 4.
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[Answer to this is 4 option 3rd.2x(x-2)+3y(y-4)+18
for this to be minimum both expressions individually should be minimu.
cannot take x or y as negative because it will reduce net expressions
in to positive value.x and y respectively cannot be greater than or equal to 2 or 3.
this gives as x as 1
and y as 2]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[/quote]
for this to be minimum both expressions individually should be minimu.
cannot take x or y as negative because it will reduce net expressions
in to positive value.x and y respectively cannot be greater than or equal to 2 or 3.
this gives as x as 1
and y as 2]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[/quote]
I guess many posts are trying to avoid it, because it seems "artificial", but it is simply a "completing the squares" traditional method.
Let me show you how you could find his "simplified" (equivalent) expression without "trial-and-error"...
2(x^2 - 2x) + 3(y^2 - 4y) + 18 = 2(x^2 - 2x +1) -2 + 3(y^2 - 4y +4) - 12 + 18 = 2(x-1)^2 + 3 (y-2)^2 -2-12+18 and the next passage is the one Rahul´s found! Simple as that.
I dont understand how did you get 2(x^2 - 2x +1) -2 + 3(y^2 - 4y +4)- 12 + 18 = 2(x-1)^2 + 3 (y-2)^2 -2-12+18
I got this:
2(x^2-2x+1)-2+3(y^2-4y+2) -6+18=2(x-1)^2+3(y-2)^2=-10
my answer is E!!! what Im doing wrong. How did you guys get 3(y^2-4y+4) according to the formula its (a^2-2ab+b^2)= (a-b)^2 so then 3(y^2-4y+2)=2(y-2)^2
Let me show you how you could find his "simplified" (equivalent) expression without "trial-and-error"...
2(x^2 - 2x) + 3(y^2 - 4y) + 18 = 2(x^2 - 2x +1) -2 + 3(y^2 - 4y +4) - 12 + 18 = 2(x-1)^2 + 3 (y-2)^2 -2-12+18 and the next passage is the one Rahul´s found! Simple as that.
I dont understand how did you get 2(x^2 - 2x +1) -2 + 3(y^2 - 4y +4)- 12 + 18 = 2(x-1)^2 + 3 (y-2)^2 -2-12+18
I got this:
2(x^2-2x+1)-2+3(y^2-4y+2) -6+18=2(x-1)^2+3(y-2)^2=-10
my answer is E!!! what Im doing wrong. How did you guys get 3(y^2-4y+4) according to the formula its (a^2-2ab+b^2)= (a-b)^2 so then 3(y^2-4y+2)=2(y-2)^2
- Shalabh's Quants
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Hi All,
I agree with Fabio.Taking assumptions... Differentiating w.r.t. X first implying Y is constant and visa-versa.In fact it is flawed partial differentiation.
Even if one intends to apply differential calculus, still there is a flaw. One needs to double differentiate to know if said function will attain maximum/minimum value.
The expression is f(x,y)=2x^2 +3y^2-4x-12y+18.
If we differentiate with respect x and equate to 0, we get
f'=4x-4 = 0
i.e. x = 1;
If we differentiate with respect y and equate to 0,we get
f'=6y-12 = 0
i.e. y = 2
when substituted x = 1, y = 2, we get expression value as 4. This value of 4 can be maximum/minimum value. To conclude, differentiate again..
So, f''=df'/dx= + 4; As f'' is +ive, it means at x=1; f(x,y) will yield minimum value and visa-versa for -ive.
Similarly, f''=df'/dy= + 6;As f'' is +ive, it means at y=2, f(x,y) will yield minimum value.
--Shalabh
I agree with Fabio.Taking assumptions... Differentiating w.r.t. X first implying Y is constant and visa-versa.In fact it is flawed partial differentiation.
Even if one intends to apply differential calculus, still there is a flaw. One needs to double differentiate to know if said function will attain maximum/minimum value.
The expression is f(x,y)=2x^2 +3y^2-4x-12y+18.
If we differentiate with respect x and equate to 0, we get
f'=4x-4 = 0
i.e. x = 1;
If we differentiate with respect y and equate to 0,we get
f'=6y-12 = 0
i.e. y = 2
when substituted x = 1, y = 2, we get expression value as 4. This value of 4 can be maximum/minimum value. To conclude, differentiate again..
So, f''=df'/dx= + 4; As f'' is +ive, it means at x=1; f(x,y) will yield minimum value and visa-versa for -ive.
Similarly, f''=df'/dy= + 6;As f'' is +ive, it means at y=2, f(x,y) will yield minimum value.
--Shalabh
Shalabh Jain,
e-GMAT Instructor
e-GMAT Instructor
- ronnie1985
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Convert the given expression into sum of squares
The exp becomes:
2(x-1)^2+3(y-2)^2+4
Min = 4
The exp becomes:
2(x-1)^2+3(y-2)^2+4
Min = 4
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min value 4
minimise (2x2 - 4x)
=> quadratic equation hence minimum value at x = 1( mid point of two roots 0 and 2), value = -2
and
minimise (3y2-12y)
similarly minimum value at y = 2 (mid point of two roots 0 and 4), => value = - 12
minimum of the function = -2 -12+18 = 4
minimise (2x2 - 4x)
=> quadratic equation hence minimum value at x = 1( mid point of two roots 0 and 2), value = -2
and
minimise (3y2-12y)
similarly minimum value at y = 2 (mid point of two roots 0 and 4), => value = - 12
minimum of the function = -2 -12+18 = 4