Problem Solving

Hi,

There are few problems for Maximum/Minimum value deduction that are tricky and time consuming. Pl. see below few examples of those with Take Away approach

Q.What is the Maximum value of =(7-x)^5 (7+x)^4?

A.7^9
B.14^4
C.0
D.7^9.2^17.5^5/3^18
E.7^9.2^7.5^7/3^18

------------------------------------------------

Take Away...

If a,b, c are variables and a+b+c is constant,

& Let X=a^p.b^q.c^r

then for x to be maximum...a/p=b/q=c/r

In above problem, say x=(7-x)^5 (7+x)^4, where a=(7-x); b=(7+x); p=5; q=4.

As a+b=(7-x)+(7+x)=14 (Constant), hence to get max of X...

=> Do (7-x)/5=(7+x)/4

=>which yields...x=-7/9

=> by putting the value of x=-7/9 in (7-x)^5 (7+x)^4, we get 7^9.2^17.5^5/3^18.

Answer is D.

Q.What is the minimum value of x+1/x ? If x>0.

A.2
B.1
C.1/2
D.sqrt 2
E.3/2
--------------------------------------------
Take Away..

If A*B=constant; where A & B are variables,
then A+B=Minimum, when A=B.


Since in given problem x+1/x,

x*1/x= 1 (Constant), hence x+1/x will be minimum when x= 1/x

=> so, x= 1/x => x^2=1 => x= 1, -1.

=> Take x=1 only as x>0.

Hence Minimum value of x+1//x=2.

Q.What is the maximum value of xy ? If 3x+4Y=24.

A.24
B.12
C.120
D.0
E.infinity
--------------------------------------------
Take Away..

If A+B=constant; where A & B are variables,
then A*B=maximum, when A=B.

We can write xy as 3x.4y/12.

As 3x+4Y=24,

so for 3x.4y to be max 3x=4y=24/2=12.

This gives x=4, y=3

The maximum value of xy would be 4.3=12.

Q.What is the minimum value of 1/x+1/y+1/z ? If x+y+z=1.

A.1
B.3
C.6
D.8
E.9
--------------------------------------------
Take Away..

If A+B+C=constant; where A, B & C are variables,
then A*B*C=maximum, when A=B=C.

=> 1/x+1/y+1/z = (xy+yz+xz)/xyz;

=> for (xy+yz+xz)/xyz to be minimum; denominator xyz should be maximum.

=> since x+y+z=1, hence at x=y=z, xyz will be max.

=> this gives x=y=z=1/3,

or, 1/x+1/y+1/z= 9.

Q. What is the minimum possible value of f(x)=max {(3-x), (2+x), (5-x)} ? Where X is an integer.

A.5
B.6
C.0
D.3
E.4

---------------------------------------

We need to get Minimum of [f(x)= max {(3-x), (2+x), (5-x)}]

Check for 0, +ive and -ive values of x and see the behaviour of fn.

By Putting x=0, f(0)= max {(3-0), (2+0), (5-0)}= max {3, 2, 5} = 5.

Similarly...

=> x=0, f(x)=5,
=> x=-1, f(x)=6,
=> x=1, f(x)=4,
=> x=2, f(x)=4,
=> x=3, f(x)=5,

As we take higher +ive values, f(x) increase so is the case with -ive values.

Hence f(x) will have minimum value as 4.

Shalabh's Quants wrote:Q.What is the maximum value of xy ? If 3x+4Y=24.

A.24
B.12
C.120
D.0
E.infinity
--------------------------------------------
Take Away..

If A+B=constant; where A & B are variables,
then A*B=maximum, when A=B.

We can write xy as 3x.4y/12.

As 3x+4Y=24,

so for 3x.4y to be max 3x=4y=24/2=12.

This gives x=4, y=3

The maximum value of xy would be 4.3=12.

Can anyone explain...Why to maximise 3x.4y? We want to maximise xy.

hi Shalabh

Good tips. Can you tell us where did you get these questions from? Are these from OG?

Thanks

iwillsurvive101 wrote:hi Shalabh

Good tips. Can you tell us where did you get these questions from? Are these from OG?

Thanks

Thanks! These are compilations from various GMAT TestPrep cos..Not from OG though.

sureshs wrote:

Shalabh's Quants wrote:Q.What is the maximum value of xy ? If 3x+4Y=24.

A.24
B.12
C.120
D.0
E.infinity
--------------------------------------------
Take Away..

If A+B=constant; where A & B are variables,
then A*B=maximum, when A=B.

We can write xy as 3x.4y/12.

As 3x+4Y=24,

so for 3x.4y to be max 3x=4y=24/2=12.

This gives x=4, y=3

The maximum value of xy would be 4.3=12.

Can anyone explain...Why to maximise 3x.4y? We want to maximise xy.

Since given is 3x+4Y=24 = constant. We can maximise the product of elements present in 3x+4Y=24 only i.e. 3x and 4y. By way of maximising 3x.4y = 12xy, we can also deduce maximim value of xy.

Shalabh's Quants wrote:

sureshs wrote:

Shalabh's Quants wrote:Q.What is the maximum value of xy ? If 3x+4Y=24.

A.24
B.12
C.120
D.0
E.infinity
--------------------------------------------
Take Away..

If A+B=constant; where A & B are variables,
then A*B=maximum, when A=B.

We can write xy as 3x.4y/12.

As 3x+4Y=24,

so for 3x.4y to be max 3x=4y=24/2=12.

This gives x=4, y=3

The maximum value of xy would be 4.3=12.

Can anyone explain...Why to maximise 3x.4y? We want to maximise xy.

Since given is 3x+4Y=24 = constant. We can maximise the product of elements present in 3x+4Y=24 only i.e. 3x and 4y. By way of maximising 3x.4y = 12xy, we can also deduce maximim value of xy.

Ok Ok. I got it sir. thank you v. much.

Hi Shalabh

Can You also explain where all we can use this concept in detail

regards
Vegi