# Find Minimum + Maximum of function with constraints

4,477

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 $${\small{\frac{3}{7}}}(4-\sqrt{2})\approx 1.108194187$$

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### Eyal Klemm

Updated on December 11, 2022

• Eyal Klemm 11 months

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 .

• Matti P. over 4 years
Judging by the tags, you are aware of Lagrange multipliers. How would you use them in this case? Did you try something?
• Eyal Klemm over 4 years
hi, first i tried to isolate x from the second constraint and then to put it in the first one , then tried to use lagarange multipliers but stucked with really ugly equations so i thought there is might be more elegant way of doing it.
• lab bhattacharjee over 4 years
$$1=x^2+y^2+z^2=(2y+3z)^2+y^2+z^2=5y^2+10z^2+12yz=10\left(z+\dfrac35y\right)^2+y^2\left(5-\dfrac{18}5\right)$$ WLOG $\sqrt{10}\left(z+\dfrac35y\right)=\cos t,\sqrt{\dfrac75}y=\sin t$ $$x^2+2y^2+3z^2=1+y^2+2z^2=?$$
• Matti P. over 4 years
The traditional approach is to take the constraint functions $g_i(x)$, multiply them by $\lambda_i$ and then add to the function that you're trying to maximise/minimise. Did you try that?
• Eyal Klemm over 4 years
yes, i tried that but ended up with a ''messy'' equations.
• A.Γ. over 4 years
Elimination a variable is apparently not a point with this exercise, it is supposed to use Lagrange multipliers in a pretty straightforward manner.
• Eyal Klemm over 4 years
Thank you for the whole explanation!
• quasi over 4 years
@Eyal Klemm: You're welcome! But I agree with your initial assessment -- solving the system is cumbersome.
• quasi over 4 years
@Eyal Klemm: Is it an exercise from a book? If so, which one?
• Eyal Klemm over 4 years
I dont know if its from a book . We got it from previous semester exam for practicing.
• quasi over 4 years
@Eyal Klemm: What textbook are you using for the course?