Exercise XV num 6 from Silvanus Thompson - Calculus Made Easy

Question

Exercise asks to verify that the sum of three quantities x, y, z, whose product is a constant k, is minimum when these three quantities are equal.

This is my amateurish attempt:

1. $x + y + z = S$;
2. $x*y*z=k$;
3. $z = \frac{k}{xy}$;
4. x + y + $\frac{k}{xy}$ = S;
5. $\frac{\partial S}{\partial x} = 1 - \frac{k}{x^2y} = \frac{x^2y-k}{x^2y}$ and symmetricly $\frac{\partial S}{\partial y} = \frac{y^2x-k}{y^2x}$;
6. $x^2y-k = 0 \rightarrow x^2y=k \rightarrow x^2y=xyz \rightarrow \frac{x^2y}{xy}=z \rightarrow x=z$ and symmetricly $y=z$, so $y=x=z$;
7. This part is doubtful to me. Proving that S is minimal when $x=y=z \rightarrow$ $\frac{\partial ^2 S}{\partial x^2} = \frac{2k}{yx^3} \rightarrow \frac{2(xyz)}{yx^3} \rightarrow \frac{2z}{x^2} \rightarrow \frac{2x}{x^2} \rightarrow \frac{2}{x}$ and symmetricly $\frac{2}{y}$;

Conclusion - if second derivative is positive, that means $S$ must be a minimum. Please be free to comment if this solution is flawed or missing something.

Answer 1

I have been battling with the same exercise for a couples of days. Can't say I have arrived at a satisfactory solution, but I will explain you what I got, which was different from yours.

Steps 1 to 5 are the same.

6. The sum of the two partial derivatives, and therefore, $\delta$S, is $\frac{(yx^2-k)}{yx^2}+\frac{(y^2x-k)}{y^2x}$;
7. Assuming x = y = z, then $\delta$S may be rewritten in terms of just x as $\frac{(x^3-k)}{x^3}+\frac{(x^3-k)}{x^3}$, which is equal to $\frac{2(x^3-k)}{x^3}$;
8. It is easier now to take the second derivative of S in terms of x, which simplifies very easily to $\frac{6k}{x^4}$;
9. x$\neq$0, because otherwise z would be undefined, as per step 3.
10. Now comes the issue. If x = y = z is a number $\gt$ 0, then the second derivative is always positive, and got our minimum value(s) for the function S. But if If x = y = z is a number $\lt$ 0, then the numerator of the second derivative is negative and its denominator positive, and the second derivative is always negative.

This if what I got. I asume either there's something wrong or that it only 'makes sense' to speak of the function for positive real numbers.

Answer 2

Constrained optimization problem:

$$\text{minimize}~~ f(x,y,z)=x+y+z \\ \text{subject to}~~~g(x,y,z)=xyz-k=0$$

Proceeding with the method of Lagrange multipliers ($\nabla f=\lambda \nabla g$) leads us to the equations $$\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}=\begin{bmatrix}\lambda yz \\ \lambda xz \\ \lambda xy\end{bmatrix}\implies yz=xz=xy=\frac{1}{\lambda}$$ But of course $yz=xz\implies x=y$, and $xz=xy\implies z=y$ and the transitive property implies $x=z$. Hence $x=y=z$.

(The same argument holds if we instead want to maximize $f(x,y,z)$.)