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1

It's true that some of the $x_i$ will tend to $\infty$ as $P$ does, however, it's not always true that all of them will. Intuitively, because the functions are increasing, the solutions must always lie on the top right-hand face of the simplex $$\Sigma_P:=\left\{x:\sum_i x_i=P\right\}.$$ A quick argument for this goes like suppose that a solution $y$ does ...


1

I didn't really see any error in your computation. However you shouldn't drop one constraint. Instead you should keep the two simpler constraints $$g_2(x,y,z)=x^2+y^2-1\\ g(x,y,z)=xy+z^2$$ Use two parameters for Lagrange multiplier: $$L=f(x,y,z)+\lambda g_2(x,y,z)+\mu g(x,y,z)$$ Then set up: $$\nabla f+\lambda \nabla g_2 +\nabla g=0$$ together with the ...


0

To continue from Daniel's answer, add the first three equations together to get: $$(2-\lambda)(4x+3y+5z)=0$$ This gives two cases: $$\lambda=0\\ 4x+3y+5z=0$$ Using $\lambda=0$, we can easily solve the others: $x=\pm 1, y=\pm 1, z=\pm 1$. This gives a minimum. The condition $4x+3y+5z=0$ will lead to contradiction from the first three equations.


1

$x$ is an extremum along $g^{-1}\{0\}\cap S$, not an extremum on $S$. That means that the gradient of $f$ doesn't necessarily have to be $0$, it just has to be perpendicular to $g^{-1}\{0\}$ at that point. What is also perpendicular to $g^{-1}\{0\}$ is the gradient of $g$. That means the gradient of $f$ and the gradient of $g$ are parallel in the extremal ...


1

The usual Fermat theorem gives you that $Df(x)=0$ simply because you can test the function in any direction around $x$. Here you can move only along directions that belong to the tangent space at $x$ to $S \cap g^{-1}(0)$, and you will not conclude that $Df(x)$ must vanish. So to speak, in constrained optimization you lack too many direction to deduce that ...


0

Consider the directional derivative of $f(x,y)$ constrained to the surface $G(x,y)=0$ at some point (a,b). This will be equal to the projection of $\nabla f(a,b)$ onto the tangent plane of the surface $G(x,y)=0$ at (a,b) . $\nabla G(a,b)$ is the normal vector to the tangent plane. Setting $\nabla f(a,b)$ parallel to $\nabla G(a,b)$ guarantees that the ...


0

That follows is a geometric argument. We consider the variations of $f(x)$ on the variety $V$ defined as $g_1(x)=\cdots=g_k(x)=0$. The Lagrange condition in $x\in V$ is: $(\star)$ there is $(\lambda_i)_i$ s.t. $Df(x)=\sum_i\lambda_iDg_i(x)$. Note that $(\star)$ is equivalent to $\cap_i \ker(Dg_i(x))\subset\ker(Df(x))$, that is $T_xV\subset\ker(Df(x))$ where ...


1

Heuristic answer (the sort you would see in, e.g., a thermodynamics class): Let $G : \mathbb{R}^2 \to \mathbb{R}$. Consider the surface $G(x,y)=0$. Then to move from a point $(x,y)$ to an infinitely close point $(x+dx,x+dy)$ on the surface, we must have $$dG=\frac{\partial G}{\partial x} dx + \frac{\partial G}{\partial y} dy = 0.$$ So we get an implicit ...


0

Let us focus our attention on the boundary of $D$. From your system we deduce $$ (4\lambda^2-1)y=1-2\lambda, $$ and then $x=-y$. This gives us a single admissible point $P_0$ whose coordinates are $x=\frac{5}{\sqrt{2}}$ and $y=-\frac{5}{\sqrt{2}}$. At this point, $f(P_0)=-\frac{25}{2}+5 \sqrt{2} \approx -5.42893$. Now we must remember that the case ...


4

Beginning with the equation for the Lagrange Multipliers, it is a matter of (tedious) algebraic manipulation with a goal to systematically eliminate parameters. We have $x+y-z=0$ and $\frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25}=1$, $f(x,y,z)=x^2+y^2+z^2$ $$\begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} = \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} + ...


1

You have two equations in three unknowns, so just have to choose one variable to maximize over. When you eliminate $z$ you have to do it from the second constraint as well as the objective function. Your problem becomes to maximize $2x^2+2y^2+2xy$ subject to $x^2/4+y^2/5+(x+y)^2/25=1=\frac {29}{100}x^2+\frac 6{25}y^2+\frac 2{25}xy$ Now solve the second ...


1

An alternative approach, if $A$ is invertible: First consider the case when $A=I$. Then the problem is to maximize $u^Tx$ subject to $x^T x\le r^2$, where $u$ is a fixed vector ($e_\alpha$ in your case). By Cauchy-Schwarz, the solution is $x=ru/|u|$. For general invertible $A$, the problem is to maximize $u^Tx$ subject to $x^TA^TAx\le r^2$. Let $y=Ax$. ...


1

Note that $x^T A^T A x = (Ax)^T (Ax) = \lVert Ax \rVert^2$, so this is expressible as a norm. We expect the derivative to be something like $2A^T Ax$, which is what you have in the line $$ \frac{\partial L}{\partial x_l} = \delta_{\alpha l} + \lambda a_{jl}(a_{ji} x_i + a_{jk}x_k): $$ changing the indices, this is $$ \frac{\partial L}{\partial x_l} = ...


6

We have $$\frac{x^2}4+\frac{y^2}9=1$$ so, any point can be written as $(2\cos t,3\sin t)$ If the distance if $d,$ $$d^2=(2\cos t-5)^2+(3\sin t-5)^2$$ Now use Second derivative test


0

As $x_1,x_2 \leq 1$ you have $V(x)=4-x_1-x_2$ on the constraint. Then according the lagrange multipliers you have a critical point \begin{align*} x_1^2+x_2^2=1\\ -1=\lambda x_1\\ -1=\lambda x_2\\ \end{align*} which has solution $x_1=x_2=\frac 1 {\sqrt 2}$ at which $V=4-\sqrt{2}$ . Compare this to the value at the endpoints of the constraint at $\left ( ...


0

Using copper.hat's hint: $x=\sqrt[p]{a^p\over ab}\\ y=\sqrt[q]{b^q\over ab}$ Show $xy=1$: $$ \begin{align} xy&=\sqrt[p]{a^p\over ab}\sqrt[q]{b^q\over ab}\\ &={a\over (ab)^{{1\over p}}}{b\over (ab)^{{1\over q}}}\\ &={ab\over (ab)^{{1\over p} + {1\over q}}}\\ &={ab\over (ab)^1}\\ &=1 \end{align} $$ From (1) above we now have: $$ ...


1

First thing first: you can simplify that $r$ out of the problem since it is homogeneous (but I would keep the $5$ since it does simplify the problem). Second thing: $f(x) = \ln x + \ln y + 3 \ln z = \ln (xyz^3)$, and $\ln$ is monotonic, so you should discard it. As Winther said, this is the only stationary point. Your octant is border-less and therefore ...


0

What requires linear constraints is the proof. The assertion the only displacements that are relevant are those that lie in the feasible set is easily true if the constraints are linear, that is, the feasible set is a linear subspace and (17.5) just says $\xi$ is parallel to the set. If the set has some curvature (degree of constraints $>1$), then (17.5) ...


0

I think lambda represents the marginal utility of the constraint. Suppose the gpa is function of studying maths and economics g(m,e), and the constraint is time (t) i.e. m+e=t. Now L=g(m,e)+λ(t-m-e). In this case, λ will represent the marginal utility of t, ie additional points in gpa that could be gained by increasing t by an hour.


0

A clever way to solve this kind of problems (with a Cobb-Douglas function) is as follows: $ax_1^{a-1}x_2^{1-a} - \lambda p_1 = 0$ $x_1^a(1-a)x_2^{-a} - \lambda p_2 = 0$ Bringing the terms involving $\lambda$ to the RHS: $ax_1^{a-1}x_2^{1-a} = \lambda p_1 $ $x_1^a(1-a)x_2^{-a} = \lambda p_2 $ Dividing the first equation by the second equation: ...


0

You should try taking the logarithm of $u(x_1,x_2)$. This is allowed because logarithms are monotonic transformations and preserve the preference relation on the set of consumption bundle. Use this to rewrite your utility function as a new function $v(x_1,x_2)$ and solve the Lagrangian multiplier problem accordingly.


0

For $$\vec a = (3, - 2,1),\vec b = \frac{1}{{\left\| {\vec a} \right\|}}\vec a$$ is given: $$f(\vec x) = \vec a \cdot \vec x$$ and constraint condition $$g(\vec x) = {\left\| {\vec x} \right\|^2} - {R^2}$$ We build the Lagrange function: $$L(\vec x,\lambda ) = f(\vec x) + \lambda \cdot g(\vec x)$$ and calculate: $${\nabla L = \left( {\begin{array}{*{20}{c}} ...


1

The distance from $(x,y,z)$ to the origin is $\sqrt{x^2+y^2+z^2}$. We therefore want to minimize $\sqrt{x^2+y^2+z^2}$, or equivalently $x^2+y^2+z^2$, subject to the condition $x^2+2y^2-z^2-1=0$. So $x^2+y^2+z^2$ should be your $f$. Now the usual process should work well.


1

You have to make sure when solving the equations that you are not dividing by something that is zero. Having done your substitution for $\lambda$, you are left with $$ y-x^2=0 \\ 2x=2xy, $$ and if $x=0$, the second equation is satisfied, and then the first equation gives $y=0$.


0

By subtracting the first constraint from the expression for the squared distance from the origin results in $-8z^2+25$ as equivalent to the function to be minimized, given the constraints. Thus the maximum occurs for $z=0$, and the maximum distance from the origin is five. This is almost obvious, because the first constraint defines an oblate spheroid ...


0

You want to minimise the distance squared from the origin, $x^2+y^2+z^2$, subject to the two conditions $ x^2+y^2+9z^2-25=0 $ and $x+3y-2z=0$. Using Lagrange multipliers means finding the stationary points of $$ L = x^2+y^2+z^2 -\lambda(x^2+y^2+9z^2-25) -\mu(x+3y-2z), $$ considered as a function of $(x,y,z,\lambda,\mu)$, and then determining which of the ...


1

Draw a picture. For either method, we want to minimize $(x-3)^2+y^2$, subject to $y=x^2$. Substituting $x^2$ for $y$ in $(x-3)^2+y^2$, we find that we want to minimize $f(x)=(x-3)^2+x^4$. Note that $f'(x)=2(x-3)+4x^3$. The critical points of $f(x)$ are where $2(x-3)+4x^3=0$, or equivalently $$2x^3+x-3=0.$$ In general cubic equations are unpleasant to ...


1

hint: $\lambda \neq 1$ since if it were you would get $2x=2x-2$ would be absurd. Thus the last equation implies $z = 0$, and the first two equations give: $\dfrac{2x}{2x-2} = \dfrac{2y}{2y-4}$. Can you solve for $x,y,z$ using the two new equations, and the constraint?


2

$x+y+z=0$ means $x,y,z$ have different signs,but for three ones ,there are two ones have same sign. WLOG, let $xy>0 \implies xy \le \dfrac{(x+y)^2}{4}$ we have $x^2+y^2+(x+y)^2=1 \ge \dfrac{(x+y)^2}{2}+(x+y)^2 \iff (x+y)^2 \le \dfrac{2}{3}$ when $x=y=\pm \dfrac{\sqrt{6}}{6} ,(x+y)^2$ get max and $xy$ get max also. $u=-xy(x+y)$, when $x>0,y>0 ...


2

This is easier to achieve along the following lines: assume that $x,y,z$ are three real roots of a polynomial $$ p(t) = t^3 + at^2 +bt +c.$$ By Vieta's formulas we have $a=0$ and $$2b=2xy+2xz+2yz=(x+y+z)^2-(x^2+y^2+z^2)=-1,$$ so $$ p(t) = t^3-\frac{t}{2}-u $$ must have three real roots. Since $f(\xi)=\xi^3-\frac{\xi}{2}$ has the following graphics: in ...



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