Find Minimum + Maximum of function with constraints

Question

homework assignment ask to find Max/Min for $$U(x,y,z) = x^2 + 2y^2 + 3z^2$$ with these constraints:

1. $x^2 + y^2 + z^2 = 1$
2. $x + 2y + 3z = 0$

Thank you.

First i tried to isolate x from the second constraint and then to put it in the first one .

Answer 1

The goal is to find the min and max values of $$f=x^2+2y^2+3z^2$$ subject to the constraints $$ \begin{cases} x^2+y^2+z^2=1&\;\;\;(\text{eq}1)\\[4pt] x+2y+3z=0&\;\;\;(\text{eq}2)\\ \end{cases} $$ Letting \begin{align*} g&=x^2+y^2+z^2-1\\[4pt] h&=x+2y+3z\\[4pt] \end{align*} and using the method of Lagrange multipliers, it follows that the critical points $(x,y,z)$ satisfy the vector equation $$\nabla{f} = a(\nabla{g})+b(\nabla{h})$$ for some $a,b \in \mathbb{R}$.

The associated scalar equations are \begin{align*} &2x = 2ax+b&&(\text{eq}3)\\[4pt] &4y = 2ay+2b&&(\text{eq}4)\\[4pt] &6z = 2az+3b&&(\text{eq}5)\\[4pt] \end{align*} With the above $3$ equations, together with $(\text{eq}1)$ and $(\text{eq}2)$, we have a system of $5$ equations in $5$ unknowns.

If two of $x,y,z$ are zero, then by $(\text{eq}2)$, all three of $x,y,z$ sould be zero, but that would contradict $(\text{eq}1)$. Hence no two of $x,y,z$ are zero.

Next we show $a\not\in\{1,2,3\}$.

• If $a=1$, then $(\text{eq}3)$ would yield $b=0$, but then from $(\text{eq}4)$ and $(\text{eq}5)$, we would get $y=0$ and $z=0$, contradiction.$\\[4pt]$
• If $a=2$, then $(\text{eq}4)$ would yield $b=0$, but then from $(\text{eq}3)$ and $(\text{eq}5)$, we would get $x=0$ and $z=0$, contradiction.$\\[4pt]$
• If $a=3$, then $(\text{eq}5)$ would yield $b=0$, but then from $(\text{eq}3)$ and $(\text{eq}4)$, we would get $x=0$ and $y=0$, contradiction.

Hence $a\not\in\{1,2,3\}$, as claimed.

If $b=0$, then since $a\not\in\{1,2,3\}$, $(\text{eq}3)$ would yield $x=0$, $(\text{eq}4)$ would yield $y=0$, and $(\text{eq}5)$ would yield $z=0$, contradiction. Hence $b\ne 0$.

Solving $(\text{eq}3)$ for $x$, $(\text{eq}4)$ for $y$ and $(\text{eq}5)$ for $z$, we get \begin{cases} x={\Large{\frac{b}{2(1-a)}}}\\[4pt] y={\Large{\frac{b}{2-a}}}\\[4pt] z={\Large{\frac{3b}{2(3-a)}}}\\ \end{cases} Replacing $x,y,z$ in $(\text{eq}2)$, we get $$\frac{b(7a^2-24a+18)}{(1-a)(2-a)(3-a)}=0$$ hence $$7a^2-24a+18=0$$ which solves as $$ a={\small{\frac{3}{7}}}(4+s) $$ where $s^2=2$ (i.e., $s=\pm\sqrt{2}$).

Replacing $a$ in $x^2,y^2,z^2$, and then rationalizing denominators, we get \begin{cases} x^2={\large{\frac{1}{4}}}b^2(43-30s)\\[4pt] y^2={\large{\frac{1}{2}}}b^2(11+6s)\\[4pt] z^2={\large{\frac{1}{4}}}b^2(11+6s)\\ \end{cases} Replacing $x^2,y^2,z^2$ in $(\text{eq}2)$, and then solving for $b^2$, we get $$ b^2={\small{\frac{1}{343}}}(19+3s) $$ Replacing $b^2$ in $x^2,y^2,z^2$, and then rationalizing denominators, we get \begin{cases} x^2={\large{\frac{1}{28}}}(13-9s)\\[4pt] y^2={\large{\frac{1}{14}}}(5+3s)\\[4pt] z^2={\large{\frac{1}{28}}}(5+3s)\\ \end{cases} Finally, replacing $x^2,y^2,z^2$ in $f$, we get $$f={\small{\frac{3}{7}}}(4+s)={\small{\frac{3}{7}}}(4\pm\sqrt{2})$$ It follows that the maximum value of $f$ subject to the given constraints is $${\small{\frac{3}{7}}}(4+\sqrt{2})\approx 2.320377241$$ and the minimum value of $f$ subject to the given constraints is $${\small{\frac{3}{7}}}(4-\sqrt{2})\approx 1.108194187$$