Posted by rohan on Thursday, June 9, 2011 at 8:58am.
Intuitively, we can locate the maximum by the following reasoning:
Given
a+2b+c=4.....(1)
as a constraint, we look for the maximum of
f(a,b,c)=ab+bc+ca...(2)
First solve for b in terms of a and b from 1 to get:
b=(4-a-c)/2
Substitute in (2) to get
(4c+4a-a²-c²)/2
which is perfectly symmetrical in a and c.
So set a and c each equal to x and find the maximum of
f(x)=8x-2x²
by setting f'(x)=8-4x=0,
we find the maximum of f(x)=f(2)=4 when a=c=2, and b=0.
Alternately, we can find it formally using the Lagrange multiplier method.
Let the objective function
P(a,b,c)=ab+bc+ca+L(a+2b+c-4)
where L is an undetermined constant, and a+2b+c-4=0 is the given constraint.
Differentiating partially with respect to a, b, and c yields the following set of linear equations:
a+2b+c=4 ....(1) given constraint
b+c+L=0 ....(2) ∂P/∂a=0
a+c+2L=0 ...(3) ∂P/∂b=0
b+a+L=0 ...(4) ∂P/∂c=0
By solving the linear system (1) to (4), we get
a=2,b=0,c=2,L=-2 as before.