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0

Even restricted, $y^*$ need not be a local maximum. Necessary would be $$ w^* \nabla_{yy} \mathcal L(y^*, \lambda^*) w \le 0 $$ for all $w^* = (w_1, w_2, w_3, 0)$ in the tangent space, that is $$ w_1 - y_3^*w_2 - y_2^* w_3 = 0. $$ Strictly less for sufficient condition. Since the (1,1) element of the hessian is $1$, it depends on $y_2^*$ and $y_3^*$.


1

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 $0$.


1

$x^2+2y^2+3z^2=x^2+y^2+z^2+y^2+2z^2=1+y^2+2z^2, x=-(y+z),(y+z)^2+y^2+z^2=1 \implies y^2+yz+z^2=\dfrac{1}{2}$, now the problem become : find min of $y^2+2z^2$ with $y^2+yz+z^2=\dfrac{1}{2}$,it shoud be easier now. or we can go further :$y=\dfrac{-z \pm \sqrt{2-3z^2}}{2},f(z)=\left(\dfrac{-z \pm \sqrt{2-3z^2}}{2}\right)^2+2z^2$ which is single variable ...


3

As Thomas commented, the function to consider is $$F=x^2+2 y^2+3 z^2+\lambda \left(x^2+y^2+z^2-1\right)+\mu (x+y+z)$$ Computing the derivatives $$F'_x=\mu +2 \lambda x+2 x$$ $$F'_y=\mu +2 \lambda y+4 y$$ $$F'_z=\mu +2 \lambda z+6 z$$ $$F'_\lambda=x^2+y^2+z^2-1$$ $$F'_\mu=x+y+z$$ From the first and second derivatives, we can eliminate $\lambda$ and ...


3

since you have $$x=\dfrac{\lambda_{2}}{2(1-\lambda_{1})},y=\dfrac{\lambda_{2}}{4-2\lambda_{1}},z=\dfrac{\lambda_{2}}{6-2\lambda_{1}}$$ and $x+y+z=0$ so we have $$\dfrac{\lambda_{2}}{2}\left(\dfrac{1}{1-\lambda_{1}}+\dfrac{1}{2-\lambda_{1}}+\dfrac{1}{3-\lambda_{1}}\right)=0$$ so we have$\lambda_{2}=0$ or ...


1

Here is how I would solve this problem using Gröbner bases in Macaulay2. R = QQ[x,y,z,l1,l2] -- define a ring I = ideal (2*x - 2*l1*x - l2, 4*y - 2*l1*y - l2, 6*z - 2*l1*z -l2, x^2+y^2+z^2-1, x+y+z) -- System of polynomial equations has isolated solutions: i7 : dim I o7 = 0 i8 : degree I -- How many are there? o8 = 4 -- We compute them ...


0

$\color{red}{Partial Answer}$ Eliminating equality constraints Define following matrices for equality constraints \begin{align} \mathbf{A}_{2\times(n+m)} &= \left[ \begin{array}{ccc} \mathbf{e_x} \space\space \mathbf{0}_{1\times n}\\ \mathbf{0}_{1\times m} \space\space \mathbf{e_y} \end{array} \right], \space \mathbf{z}_{(n+m)\times 1} = \left[ ...


1

It will perhaps clarify the situation if we make a geometric interpretation of the problem. The function $ \ f(x,y,z) \ = \ x^2 + y^2 + z^2 \ $ is the "distance-squared" function in $ \ \mathbb{R}^3 \ $ for distances of points from the origin. The "constraint surface" is a triaxial ellipsoid centered on the origin with its own axes aligned with the ...


1

The distribution maximizes $Var(X)$ is: $Pr(X=1)=\mu, Pr(X=0)=1-\mu$. Intuitively, you want to pull the distribution to the sides as much as possible under the constraint of $E(X)=\mu$.


2

If you remove the triangle with corners (0,1),(1,0),(0,0),and rotate by $\pi/4$ the problem is identical to Dido's,in variational calculus in standard text books.


0

Because of the peculiar character of this function, we can make the substitution $ \ u \ = \ x + y \ $ to reduce the problem to finding the extrema of a function of a single variable, $ \ u^2 \ - \ u \ + \ 1 \ $. The domain to be studied becomes $ \ 0 \ \le \ u \ \le \ 2 \ $ . In the "interior" of the interval, we find, as godonichia remarks, that $ \ ...


0

As I said in the comments, I would suggest you should not insist on finding a closed-form solution to problems like this. There is nothing wrong with the notion that a model is only solvable numerically. Indeed, few actually interesting models are! It looks to me like you've already tried to simplify this, again unnecessarily, before you even posted about ...


0

As has been remarked in the comments, the Lagrange-multiplier method only looks for level lines, planes, surfaces etc. which are tangent to the "constraint" curve, surface, etc. The constraint $ \ \sqrt{x} \ + \ \sqrt{y} \ = \ 4 \ $ is the portion of the astroid curve (marked in blue) in the first quadrant; "Lagrange" gives you the level curve $ \ f(x) \ = ...


-4

HINT: you can also write $f(x,(4-\sqrt{x})^2)=3x+(4-\sqrt{x})^2$


1

minimize $(x-2)^2 + y^2$ subject to $y^2 - x^2 = 4$ using Lagrange multipliers you will end up with: $2(x-2) + \lambda(-2x) = 0$ $2y + \lambda(2y) = 0$ $y^2 - x^2 = 4$ you get $\lambda = -1$, $x=1$, and $y=\sqrt{5},-\sqrt{5}$


1

HINT: use $d=\sqrt{(2-x)^2+4+x^2}$


3

I do not think you need LaGrange multiplier. Define $f(x)$ to be the square of the distance between the point $(2,0)$ and the hyperbola. (It is enough to minimize the square of the distance to make the calculus easier). Using the top half of the hyperbola, with a bit of math we get $$f(x)=(x-2)^2+(\sqrt{x^2+4}-0)^2.$$ Minimize this function using the ...


0

In general, it's not possible getting rid of the inequalities, or the Lagrange multipliers. In other words, there isn't a system of equations you can solve. If you wanna do this so badly, you can always compute a zero of the gradient, and if it doesn't satisfy the inequalities, try to find the minimum in your boundary.


0

There is no need to use Lagrange multipliers, because your constraint explicitly gives $y$ in terms of $x$. Instead, work out the range of possible $x$ values. You have $x \geq 0$ already. If you are going to have $y \geq -2$ then you have $4-2x \geq -2$, or $x \leq 3$. So you want to optimize the one variable function $g(x)=f(x,4-2x)$ over $[0,3]$.


1

You can get rid of one constraint for free: All three variables $x$, $y$, $z$ have to be positive. It is therefore allowed to put $$x:=u^2,\quad y:=uv,\quad z:=v^2\ ,$$ so that $xz=y^2$ is automatically satisfied. We now have to investigate the function $$g(u,v):=(2a+b)\log u+(2c+b)\log v$$ under the sole constraint $u^2+uv+v^2=1$. We obtain the conditions ...


0

I used KittyL's suggestion to isolate x and y:


0

You made some mistakes in method 1. Equation (6) should be $2x+2y=2\lambda+\mu(2x+2y)$ Equation (7) has a similar mistake. I didn't understand what you really did when you do "(6) in (7)". Method 1 seems not work. Try this: Transform equation (1),(2),(3): $$x=\frac{\lambda}{2(1-\mu)}\\ y=\frac{\lambda}{2(1-\mu)}\\ z=\frac{2\lambda-\mu}{2}$$ You should ...


0

Hint There is another way which could be the minimization of the square of the distance and the elimination of one variable from the equality constraint. So, eliminate $z$ from the constraint $$z=14-2x-3y$$ and plug it into $x^2+y^2+z^2$. So, you want to minimize $$F=x^2+y^2+(14-2x-3y)^2$$ Compute the partials $$F'_x=10 x+12 y-56$$ $$F'_y=12 x+20 y-84$$ Say ...


0

If you want to do it with Lagrange multipliers, it's better to consider $d^2$, since $d$ is always positive. But that's not the only way in this case: because the constraint is that the point lie in a plane, you can recall that the shortest distance vector will be perpendicular to the plane; but we know what this is: a plane is given in the form $$ {\bf n} ...


0

Then Let g(x,y,z)=2x+3y+z-14=0 find$$f_{x}=L*g_{x}$$ $$f_{y}=L*g_{y}$$ $$f_{z}=L*g_{z}$$ Where $f(x,y,z)=x^{2}+y^{2}+z^{2}$ You have four simultaneous equation with four unknowns and you can thus solve the equation. You also have another method $1)$Replace the constraint equation in the equation to be maximized or minimized $2)$Then you have a new ...


1

Hint Let $d^2(x,y,z)=f(x,y,z)=x^2+y^2+z^2$ and $g(x,y,z)=2x+3y+z-14=0$ and use Lagrange multipliers as follow $$\begin{cases} f_x=\lambda g_x \\ f_y=\lambda g_y\\ f_z=\lambda g_z\\ g(x,y,z)=0 \end{cases} $$ After some algebra I got $$x=2, y=3, z=1$$



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