# Tag Info

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 ...

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 ...

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.

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$x^2+2y^2+3z^2=x^2+y^2+z^2+y^2+2z^2=1+y^2+2z^2, x=-(y+z),(y+z)^2+y^2+z^2=1 \implies y^2+yz+z^2=\dfrac{1}{2}$, 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 ...

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 ...

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 ...

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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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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 $y=\sqrt{5},-\sqrt{5}$

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

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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 $E(X)=\mu$.

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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 ...

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