## Monday Math 151

Find the values of x,y,z≥0 that maximize the value of f(x,y,z)=xyz, subject to the constraint ax+by+cz=d, where a, b, c and d are positive real numbers.

This is a maximization subject to constraint problem, and is thus best approached by the method of Lagrange multipliers. Thus, we find the stationary points of the function:
$F(x,y,z,\lambda)=xyz-\lambda(ax+by+cz-d)$
(where we have chosen –λ rather than +λ to make some later algebra easier).
Taking the partial derivatives of F, we see
$\frac{\partial{F}}{\partial{x}}=yz-\lambda{a}$
$\frac{\partial{F}}{\partial{y}}=xz-\lambda{b}$
$\frac{\partial{F}}{\partial{z}}=xy-\lambda{x}$
$\frac{\partial{F}}{\partial\lambda}=-(ax+by+cz-d)$
Setting the first three of these equal to zero, we obtain
$yz=\lambda{a}$
$xz=\lambda{b}$
$xy=\lambda{c}$
And setting the fourth equal to zero gives us our constraint $ax+by+cz=d$, as expected.
Now, multiplying those three equations we obtained from setting the partial derivatives equal to zero, we obtain
$x^2y^2z^2=\lambda^3{abc}$
and taking the square root of both sides (with λ≥0),
$xyz=\sqrt{\lambda^3{abc}}$
Dividing this by the first equation, we see
$\begin{array}{rcl}\frac{xyz}{yz}&=&\frac{\sqrt{\lambda^3{abc}}}{\lambda{a}}\\{x}&=&\frac{\sqrt{\lambda{abc}}}{a}\end{array}$
And analogously,
$y=\frac{\sqrt{\lambda{abc}}}{b}$
$z=\frac{\sqrt{\lambda{abc}}}{c}$
And thus, $ax=by=cz=\sqrt{\lambda{abc}}$
and from our constraint $ax+by+cz=d$, we see that
$ax=by=cz=\frac{d}{3}$
(from which we can solve for λ, but this is unnecessary).
Thus, the point which maximizes f(x,y,z)=xyz is $(x,y,z)=\left(\frac{d}{3a},\frac{d}{3b},\frac{d}{3c}\right)$, with the maximum value thus being $\frac{d^3}{27abc}$.