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}.

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