Question

Find the greatest value of `abc` for positive values of `a,b,c` subject to `ab +bc+ca=12` ?

Answer 1

`abc=8`

Explanation:

Here's a method using a Lagrangian...

Let:

`{ (a=e^u), (b=e^v), (c=e^w) :}`

where `u, v, w` are real numbers.

That takes care of the requirement that `a, b, c > 0`.

We want to maximise:

`f(u,v,w) = e^ue^ve^w`

subject to:

`g(u,v,w) = e^ue^v+e^ve^w+e^we^u = 12`

Define the Lagrangian:

`L(u,v,w,lambda) = f(u,v,w)-lambda(g(u,v,w)-12)`

`color(white)(L(u,v,w,lambda)) = e^ue^ve^w-lambda(e^ue^v+e^ve^w+e^we^u-12)`

At the maxima and minima we have:

`grad L(u,v,w,lambda) = bb(0)`

That is, all of the partial derivatives of `L` are zero:

`0 = (del L)/(del u) = e^u(e^ve^w-lambda (e^v+e^w))`

`0 = (del L)/(del v) = e^v(e^we^u-lambda (e^w+e^u))`

`0 = (del L)/(del w) = e^w(e^ue^v-lambda (e^u+e^v))`

`0 = (del L)/(del lambda) = -(e^ue^v+e^ve^w+e^we^u-12)`

That is:

`{ (bc-lambda(b+c) = 0), (ca-lambda(c+a) = 0), (ab-lambda(a+b)=0), (ab+bc+ca=12) :}`

We find:

`lambda = (bc)/(b+c) = (ca)/(c+a) = (ab)/(a+b)`

Hence:

`c = (lambdab)/(b-lambda) = (lambdaa)/(a-lambda)`

Hence:

`lambdaab-lambda^2b = lambdaab-lambda^2a`

Hence `a=b` and similarly `b=c`

Hence `a=b=c` and `12 = 3a^2`, so `a=2` and `abc=8`

`color(white)()`
Footnote

The Lagrangian method used here is a standard method for finding maxima and minima of continuous functions of several variables subject to constraints.

In this particular example, there may be an alternative algebraic method involving `(a-b)^2+(b-c)^2+(c-a)^2` or similar, especially since we have found that the solution occurs when `a=b=c`.