Thursday

June 30, 2016
Posted by **rohan** on Thursday, June 9, 2011 at 8:58am.

- algebra -
**MathMate**, Thursday, June 9, 2011 at 11:20amIntuitively, 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.