Lagrange Multiplier with $3$ variables

Question

Here is the question: $f(x,y,z) = x + 2y^2 - 3z$ subject to the constraint $z = 4x^2 + y^2$.

I don't understand how to do this because when I take the partial derivative in respect to $z$, I get $-3 = 0$, which doesn't make sense.

Can someone please help me? I don't need it all worked out, just what to do from here...

Answer 1

\begin{align} \max_{x,y,z}\quad &x+2y^2-3z\\ s.t. \quad 4x^2+y^2-z&=0\\ \therefore \mathcal{L}(x,y,z,\lambda)&=x+2y^2-3z-\lambda\cdot(4x^2+y^2-z)\\ \nabla\mathcal{L}(x,y,z,\lambda)&= \begin{pmatrix}1\\4y\\-3\end{pmatrix}- \lambda\cdot\begin{pmatrix}8x\\2y\\-1\end{pmatrix}=0\\ \mbox{Now, From Equation (3):}\\ \implies-3+\lambda&=0\implies \lambda=3.\\ \mbox{Now, From Equation (1):}\\ \therefore 1-3\cdot8x&=0\implies x={1\over24}\\ \mbox{Now, From Equation (2):}\\ 4y-3\cdot2y&=0\implies y=0\\ \mbox{Now, from the Equality Constraint:}\\ 4x^2+y^2-z&=0\implies z={1\over144}\\ \mbox{Solution}&=\{{1\over24},0,{1\over144}\} \end{align}

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Let's $\ds{{\cal F} \equiv x + 2y^{2} - 3z - \mu\pars{z - 4x^{2} - y^{2}}}$

$$ \begin{array}{rclcrcrcl} 0 & = & \partiald{{\cal F}}{x} & \imp & 1 & + & 8\mu x & = & 0 \\ 0 & = & \partiald{{\cal F}}{y} & \imp & 4y & + & 2\mu y & = & 0 \\ 0 & = & \partiald{{\cal F}}{z} & \imp & -3 & - & \mu & = & 0 \end{array} $$

Then, $\mu = -3$ $$ \imp\quad x = -\,{1 \over 8\mu} = {1 \over 24}\,,\quad y = 0 $$

Answer 3

f(x,y) = x + 2y^2 - 3(4x^2 + y^2) = -12x^2 + x - y^2. The function g(x) = -12x^2 + x is a parabola with vertex x = -b/2a = -1/(2(-12)) = 1/24. So max g = g(1/24) = 1/48 ==> max f = 1/48 at (x,y,z) = (1/24,0,1/144).