Lagrange Multipliers — Question 10

Question 10 – Solution

Strategy. Multiplying each component equation by its corresponding coordinate reveals fixed ratios among $x^2,y^2,z^2$.

Step 1: Nonzero-product candidates The multiplier system is $$yz=2\lambda x,\qquad xz=4\lambda y,\qquad xy=6\lambda z.$$ When $xyz\ne 0$, multiply these equations by $x,y,z$, respectively: $$xyz=2\lambda x^2=4\lambda y^2=6\lambda z^2.$$ Thus $$2x^2=4y^2=6z^2=K.$$ Substitution into the constraint gives $$\frac K2+2\frac K4+3\frac K6=\frac{3K}{2}=6,$$ so $K=4$. Therefore $$x^2=2,\qquad y^2=1,\qquad z^2=\frac 23.$$

Step 2: Values and zero cases At these eight sign choices, $$|xyz|=\sqrt 2\cdot 1\cdot\sqrt{\frac 23}=\frac 2{\sqrt 3}.$$ Any feasible candidate with a zero coordinate has $f=0$, which lies strictly between the positive and negative nonzero values.

Step 3: All extrema The ellipsoid is compact, so comparison is decisive: $$\boxed{f_{\max}=\frac 2{\sqrt 3}}$$ at the four points $(\pm\sqrt 2,\pm 1,\pm\sqrt{2/3})$ having an even number of minus signs, and $$\boxed{f_{\min}=-\frac 2{\sqrt 3}}$$ at the four such points having an odd number of minus signs. The parity rule follows because the product is positive for an even number of negative factors and negative for an odd number.