Cheeky approach to few tricky problems-3

Shalabh's Quants wrote:What is the minimum value of (ab+ac+ad+bc+bd+cd)? If abcd = 16.

A.51
B.26
C.12
D.24
E.49
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Hit & Trial Method...

simply take a=b=c=d => This gives a=b=c=d = 2 each.

Now put a=b=c=d = 2 in (ab+ac+ad+bc+bd+cd);

we get (ab+ac+ad+bc+bd+cd) as 24.

You may try with other combinations for a, b, c, d as say 1, 1, 1, 16 or something other.

In each case you will land up greater value than 24.

Algebraic approach...

To get sum of few variables to be minimum, their product should be constant & all should be equal to each other.

pl. see this https://www.beatthegmat.com/must-see-for ... 09814.html

so for (ab+ac+ad+bc+bd+cd) to be min. => ab.ac.ad.bc.bd.cd should be constant.

ab.ac.ad.bc.bd.cd = (abcd)^3 = 16^3 (constant)

Real good one! Thank you!

Only applicable when multiplication of nos. is constant.

What is visa-versa? Is there anything also if sum contant.

spider wrote:
Real good one! Thank you!

Only applicable when multiplication of nos. is constant.

What is visa-versa? Is there anything also if sum contant.

Raised a good question spider.
Yes there is a trick when the sum is constant.
When the sum of two or more numbers is constant then their multiplication will yield a maximum value when all the numbers are equal.
Observe the table
taking the sum of two nos constant = 10

1+9 = 10 | 1*9 = 9
2+8 = 10 | 2*8 = 16
3+7 = 10 | 3*7 = 21
4+6 = 10 | 4*6 = 24
5+5 = 10 | 5*5 = 25

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again,
taking the sum of three nos constant = 15

5+5+5 = 15 | 5*5*5 = 125
4+5+6 = 15 | 4*5*6 = 120
3+5+7 = 15 | 3*5*7 = 105
2+5+8 = 15 | 2*5*8 = 80
1+5+9 = 15 | 1*5*9 = 45

Take any combination maximum value will be 125.
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Cheers,
Niket

I did not understand the login...even if it is trial and error..how did we get the first trail that it should be 2...just guess?

Can you please explain in a more elaborate way ....

[email protected] wrote:I did not understand the login...even if it is trial and error..how did we get the first trail that it should be 2...just guess?

Can you please explain in a more elaborate way ....

Take Away..

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



Take Away..

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