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The "Lagrangian function" $L$ is a purely formal device without any intuitive content, but it condenses the complex geometric data into a simple recipe. The "method" in question relies on the following fact: Given a function $f:{\mathbb R}^n\to{\mathbb R}$ and a surface ("condition") $$S:\quad g(x)=c\ ,$$ the function $f$ cannot assume a conditional local ...


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Geometric Solution: If $r$ is the shortest distance from $B(0,b)$ to the parabola $P:=\big\{(x,y)\in\mathbb{R}^2\,|\,y=x^2+8\big\}$, then the circle $\Gamma$ centered at $B$ with radius $r$ is tangent to this parabola. Suppose that $A(p,q)$ is a tangential point between $\Gamma$ and $P$. Then, $q=p^2+8$ and the line $\ell$ passing through $A$ tangent to ...


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You at least need differentiability for existence of KKT. But if you focus on sufficiency, you can say this: Sufficiency: Suppose your problem is to minimize $f(x)$ over a convex set $X$ and subject to $g_k(x)\leq 0$ for all $k \in \{1, \ldots, K\}$ (call this Probelm P1). Define: $$ L(x, \lambda) = f(x) + \sum_{k=1}^K\lambda_k g_k(x) $$ Assume that ...


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I know that this question is tagged as a Lagrange multiplier question, but as it turns out, you don't need a Lagrange multiplier because the inequality ($x+y\le 24$) is not tight, so it can be ignored. The derivation is really simple. The total points are given by the equation, $$p=200+x(50-x)+y(48-y)-(x+y)^2.$$ At the maximum, $p$ is extremal with ...


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the function you are inserted in is $f(x,y) = 200 +x(50-x) +y(48-y) - (x+y)^2$take the derivative for $x$ and $y$ and set is zero. This will give you the condition for a local extremum. The answer should be $x = 26/3,\ y = 23/3$. Then you still have to explain why that this is a local maximum and not a minimum. And at last you should check the global points. ...


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A few things: First of all (and most importantly), you call this a Lagrange problem (and your solution does satisfy x + y = 24), but is it really? Seems improbable that the optimal strategy would involve no sleep. I'd look in the interior, $x + y < 24$. Second, I don't get the same numbers you get. I get the same x and y values (using Lagrange, ...


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In this case, $q_0$ is all of the "outside" goods, that is, the more it is consumed, the higher the utility will be. This is because the utility function is monotonically increasing in $q_0$, which suggests that the budget constraint should be \begin{equation} q_0 + p_1 q_1 + p_2 q_2 \le Y \end{equation} for budget level of $Y$, and prices of $p_1$ and ...


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If any of $x$, $y$, and $z$ is $0$, then two of them are $0$ and the other is not (so as to satisfy $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$). Thus, $\lambda=0$. Hence, all such solutions are $(x,y,z,\lambda)=(\pm a,0,0,0)$, $(x,y,z,\lambda)=(0,\pm b,0,0)$, and $(x,y,z,\lambda)=(0,0,\pm c,0)$. If none of $x$, $y$, and $z$ is $0$, then ...


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The equation $\frac{\partial \mathcal L (x,\lambda)}{\partial \lambda} =0 $ just says that $g(x) = 0$. This must be satisfied, of course, if $x$ is a minimizer of $f$ subject to the constraint that $g$ is $0$.


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Let's start with the following question, suppose $$ \Sigma:=\left\{x\in\Bbb R^m \mid f(x)=0\in\Bbb R^q,\quad f(x)\in\mathscr C^1(\Bbb R^m),\quad 1\le q<m \right\} $$ and there is a scalar field, or the so-called "goal function" $\theta(x)\in\Bbb R$ on $\Sigma$. What we are going to do is seek $x_{*}$ such that $$\theta ...


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As I pointed out in a comment, the problem is quite simply solved using one-variable techniques. However, let us explore your choice of letting $x$ be the ticket price and $y$ the number of tickets sold. Then the number of $5$ dollar increments is $\frac{x-50}{5}$. The resulting number $y$ of tickets sold is given by $$y=1000-10\cdot \frac{x-50}{5}.$$ ...


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I don’t think you even need to take $X,Y$ as you have said. We can define $g:\mathbb{R}^4\rightarrow \mathbb{R}^2$ as $$g(x,y,u,v)=(x^2+y^2-a^2,u^2+v^2-b^2)\equiv(g_1,g_2)$$ For the Lagrange Multipliers Theorem to apply, we need the preimage $g^{-1}(\{0\})$ to be a $2$-dimensional sub-manifold of $\mathbb{R}^2$. It is sufficient to show that ...


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This is a two-dimensional analogue of something you are probably familiar with: Suppose a differentiable function $f:\mathbb R\rightarrow \mathbb R$ is restricted to an interval $[a,b]$ and you want to find the extrema there. Then, the procedure you follow is: Find the local extrema in $(a,b)$ by checking the values of $x$ in $(a,b)$for which $f'(x)=0$ or ...


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The theory for inequality constraints, that is $g(x,y) \le 1$, is much more complicated than for equations like $g(x,y) = 1$. As start you could consult the Wikipedia on Karush–Kuhn–Tucker conditions. However, in your simple case $f,g:\mathbb R^2\to\mathbb R$, it boils down to following: You can first try to find a maximizer in the interior, that is, ...


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The constant does not matter, as only the (total) derivative of $F$ is used in the Lagrange Multipliers method, so additive constants do not affect this. You may include the $c$, or you may not. You could argue that the $F$ found on Wikipedia contains more information about the problem than that in the example, but it proves to be irrelevant to the method. ...


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Setting $f_x=\lambda g_x, \;f_y=\lambda g_y, \;f_z=\lambda g_z$ gives $\;\;\displaystyle2x=\lambda\cdot\frac{2x}{a^2}, \;\;2y=\lambda\cdot\frac{2}{b^2},\;\;2 z=\lambda\cdot\frac{2z}{c^2}$. Therefore $\textbf{1)}$ $x=0$ or $\lambda=a^2\;\;\;$$\textbf{2)}$ $y=0$ or $\lambda=b^2\;\;\;$ $\textbf{3)}$ $z=0$ or $\lambda=c^2$ Since ...



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