Find the minimum and maximum of $f(x,y,z)=3x+2y+4z$ subject to constraint $x^2+2y^2+6z^2=9$

Question

I have $g=g(x,y,z,\lambda)=3x+2y+4z-\lambda(x^2+2y^2+6z^2-9)$

$g_x=3-2x\lambda=0$, so $\lambda=\frac{3}{2x}$

$g_y=2-4y\lambda=0$, so $\lambda=\frac{1}{2y}$

$g_z=4-12z=0$, so $\lambda=\frac{1}{3z}$

From here I'm not sure what to do.

Answer 1

You were on the right track. You also need $g_\lambda=-(x^2+2y^2+6z^2-9)=0$.

Expressing $x, y, $ and $z$ in terms of $\lambda$ $\left[x=\dfrac3{2\lambda}, y=\dfrac1{2\lambda}, z=\dfrac1{3\lambda}\right]$

yields $\left(\dfrac3{2\lambda}\right)^2+2\left(\dfrac1{2\lambda}\right)^2+6\left(\dfrac1{3\lambda}\right)^2=9.$

Now solve for $\lambda$. Then you can solve for $x, y, $ and $z$.

Finally you can calculate $3x+2y+4z$.

Answer 2

An alternative method is to use the Cauchy-Schwarz inequality. The $3$-tuples version states that $$(a_1^2 +a_2^2 + a_3^2)(b_1^2 +b_2^2 + b_3^2)\ge (a_1 b_1 +a_2 b_2 + a_3 b_3)^2$$ for all real $3$-tuples $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3),$ with equality holding if and only if $a_1=a_2=a_3=0$ or there exists a real number $r$ such that $(a_1 r,a_2 r,a_3 r) = (b_1,b_2,b_3).$

Let $(x,y,z)$ be a real triple satisfying $x^2+2y^2+6z^2 = 9.$ Choosing \begin{align*} (a_1,a_2,a_3) &= \left(3,\sqrt{2},\frac{2\sqrt{2}}{\sqrt{3}}\right),\\ (b_1,b_2,b_3) &= (x,\sqrt{2}y,\sqrt{6}z), \end{align*} the Cauchy-Schwarz inequality yields \begin{align*} 123 &= \left(3^2+\sqrt{2}^2+\left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^2\right)\cdot 9\\ &= \left(3^2+\sqrt{2}^2+\left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^2\right)(x^2+2y^2+6z^2)\\ &= \left(3^2+\sqrt{2}^2+\left(\frac{2\sqrt{2}}{\sqrt{3}}\right)^2\right)(x^2+(\sqrt{2}y)^2+(\sqrt{6}z)^2)\\ &\ge (3x+2y+4z)^2. \end{align*} Thus, $\sqrt{123}\ge |3x+2y+4z|,$ which yields the bounds $$\sqrt{123}\ge 3x+2y+4z\ge -\sqrt{123}.$$ Using the equality criteria for Cauchy-Schwarz, we can derive equality cases: the upper bound is achieved at $$(x,y,z)=\left(\frac{9\sqrt{3}}{\sqrt{41}},\frac{3\sqrt{3}}{\sqrt{41}},\frac{2\sqrt{3}}{\sqrt{41}}\right)$$ and the lower bound is achieved at the negative of that vector.