# Fimd the maximum value

1. Nov 4, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
Let a,b,c be real numbers such that a+2b+c=4. Then the max value of (ab+bc+ca) is

2. Relevant equations

3. The attempt at a solution
I am trying to substitute for b into the second expression but that won't give me the answer. Any help would be appreciated.

2. Nov 4, 2012

### Staff: Mentor

Why?
The approach is fine, what do you get as result?

3. Nov 4, 2012

### AJ Bentley

Note that a + c = 4-2b
and ab + bc + ac is the same as b(a + c) + ac

So you can write X = b(4-2b) + ac.
That's a quadratic that you can differentiate wrt b to find a maximum for X at a particular b. You can also find what ac must be from the same result.

After that, a few substitutions gives you the answer.

Last edited: Nov 4, 2012
4. Nov 4, 2012

### utkarshakash

$b=\dfrac{4-a-c}{2}$
Now substituting this into the second expression and simplifying I get
$\dfrac{4(a+c)-(a^2+c^2)}{2}$

Whats next?

5. Nov 4, 2012

### utkarshakash

Here's what I did

$-2b^2+4b+ac$
Differentiating wrt b and setting it to 0
-4b+4=0
b=1
a+c=2
Now , ac=c(2-c)
Diff again wrt c and setting it to 0
c=1
a=1
Substituting a,b and c I get maximum value = 3

But the answer is 4. I can't find out my mistake. Maybe there is something wrong with differentiation.

6. Nov 4, 2012

### HallsofIvy

Staff Emeritus
The derivative with respect to a is 2- a and the derivative wth respect to c is 2- c. Set those equal to 0 and you get a= c= 2 which then gives b= 0.

7. Nov 6, 2012

Thanks