# Tag Info

1

It is usually understood that $f_0(x)=+\infty$ for $x\notin D$. This would make both definitions equivalent.

1

Since the utility function has the Leontief form, then the two goods are perfect complements. Therefore the consumer will always choose the kink point where $a_1x_1 = a_2x_2$, i.e. the maxima $(x_1^*,x_2^*)$ satises $a_1x_1^* = a_2x_2^*$. Also since the consumer will spend all his/her income, we can have two variables $(x_1^*,x_2^*)$ and two equalities: ...

2

Let $v_i=u_{i+1}-v_i$ for $i<n$, let $v_n=\Sigma u_i$, then in the new coordinates the condition becomes $v_n=0$ and $v$ lies on a sphere in that hyperplane. This converts the problem into finding the spectrum of $A$ composed with the projection on that hyperplane.

3

Without loss of generality, we can assume that $A$ is symmetric since $f$ is a homogeneous quadratic form (if $A$ is not symmetric, note that $f(u)=\frac 12u^t\cdot (A+A^t)\cdot u$). Therefore $\nabla f=2 A\cdot u$. Let $V$ denote the variety $u_1+\dots+u_n=u_1^2+\dots+u_n^2-1=0$. Note that $V$ is smooth and compact, hence the minimum of $f$ is reached at ...

1

The Lagrange multipliers schema states that if $\Omega\subset\mathbb{R}^n$ is open, and $f:\Omega\to\mathbb{R}$ is differentiable, and if $m\leq n$ and $g_1,g_2,\ldots,g_m:\Omega\to\mathbb{R}$ are continuously differentiable, then at every local extremum $a\in U$ of $f$ subject to the constraint $g_i=0$ for $i=1,2,\ldots,m$ there exist ...

0

You really do have to check if $$F(x,y,z) = 2x^{2}+xy+y^{2}+yz+z^{2}-6x-7y-8z-9$$ has a max, a min, or neither. There's no simple derivative test for 3 variables. There's a trick that can be used to get rid of the cross terms. First, let $x=x'+\alpha y$, and choose $\alpha$ so that the $x'y$ term vanishes: $$F(x'+\alpha ... 0 You want to minimize (and maximize)$$ \Vert G u \Vert^2 \ \text{ subject to } \ \Vert u \Vert^2=1. $$The Lagrangien is given by$$ \mathcal L(u,\lambda) = \Vert G u \Vert^2 - \lambda( \Vert u \Vert^2 - 1), $$and the first order condition is$$ \frac{\partial \mathcal L}{\partial u} (u, \lambda) = 2 G^\top Gu - 2 \lambda u $$meaning that ... 1 There's no need to apply method of Lagrangian multiplier. But you can still apply it by introducing a dummy variable, say t=0. Then you can write$$\mathcal L(w;t)=f(w)-\lambda t$$0 You've posed two questions. Perhaps we'll give you a solution for the first one and leave the other as an exercise? Fair? Let a, b and c be length, width and height. We write F(a,b,c)=(2+7)ab+5c(a+b)=9ab+100\frac{a+b}{ab} . Then: \\ \frac{\partial F}{\partial a} = 9b - 100/a^{2} \equiv 0 \\ \frac{\partial F}{\partial b} = 9a -100/b^{2} \equiv 0. This ... 1 The answer is that the appropriate gradient is zero. Suppose \hat{x} is a minimizer of f subject to the constraint g(x) = 0, and Dg(\hat{x}) \neq 0. Suppose in particular that Dg(\hat{x})e_1 \neq 0 (at least one component is non-zero, it simplifies notation to assume it is the first). Then the implicit function theorem gives the existence of a ... 1 The function g defines a surface(in this case a curve) over which we have to minimize the value of f. The condition simply means that the gradients should be parallel or anti-parallel, otherwise we can increase f by moving in the direction of derivative appropriately. For more read about the method of lagrange multipliers. 2 It's not f that has a local extremum at x_{1}, but rather f|_{S}. Consider, for example, f(x,y)=xy restricted to the line y=1-x. This restricted function has a local maximum at (\frac{1}{2},\frac{1}{2}), but the full function f is not at an extremum at (\frac{1}{2},\frac{1}{2}). 1 Try minimizing f(x,y) = (\frac{x^n+y^n}{2}) subject to the constraint (\frac{x+y}{2})^n = C, where C>0 is a constant. You should be able to show that when the minimum occurs, x=y, which will give you the result you want. 1 You are right that it should be {d\over dr}f(x(r)). The result will be true for any r. When you solve the Lagrange problem, you will find x=x(r) and \lambda=\lambda(r). Substitute x=x(r) into the objective function, and you get the general formula$${d\over dr}f(x(r))=-\lambda_i(r)$$1 Note that your function is unbounded. Your constraint gives x = 1+y^2, and plugging this into the objective, you get 2\,(1+y^2)^2 + y. Now, it is easy to see, that this can attain arbitrary large values. This means that your problem has no global maximum, but it may have a local one. Further, since the gradient of your constraint is (1, -2\,y), the ... 0 From y=bx you get x^2+y^2+z^2=x^2(1+b^2)+z^2 and xyz=a\iff xbxz=a\iff x^2=\frac{a}{bz}, hence determine the extrema of \frac{a}{b z}(1+b^2)+z^2. 1 To get the dual function, you want to plug in the optimal value of \mu^*:$$ D(\lambda) = \max_\mu f(\mu) - \lambda g(\mu) = \frac{a^T S a}{4 n \lambda} + C\lambda $$Then you minimize this dual function to find the optimal \lambda.$$ \frac{\partial D}{\partial \lambda} = -\frac{a^T S a}{4 n \lambda^2} + C  \lambda^* = \sqrt{\frac{a^T S a}{4 n ...

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