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15

Here's some intuition. First suppose that $x_0$ is a local minimizer for the problem \begin{align} \tag{$\spadesuit$} \text{minimize} & \quad f(x) \\ \text{subject to} &\quad Ax = b. \end{align} If $Au = 0$, then the directional derivative $D_u f(x_0) = \langle \nabla f(x_0),u \rangle$ is nonnegative. Otherwise, it would be possible to improve ...


4

Write $\alpha_i^2=\alpha_i^{\frac{1}{2}}\alpha_i^{\frac{3}{2}}$, and then use the Cauchy-Schwarz inequality.


4

Do you have to use Lagrange multipliers? The sphere is centered at the origin, and the point lies outside it. If you find the line through $(3,1,-1)$ and the origin, the closest and farthest points will be the two points of intersection between the line and the sphere. Seeing that the length of the vector from the origin to the point of interest is ...


3

One approach is to multiply the first equation by 9, the second by 5, the third by 45, giving all 3 equations a RHS of $45\lambda$, so we can set them equal to each other. $9y+27z=45\lambda$ $5x+10z=45\lambda$ $135x+90y=45\lambda$ Set the first equal to the second, the second equal to the third, to give: $9y+27z=5x+10z$ $5x+10z=135x+90y$ Now, simplify ...


3

Look at the tangent space to $S = g^{-1}(c)$ at $x_0$, say $S_{x_0}$. Then $S_{x_0} = \left[\nabla g(x_0) \right]^{\perp}$. Therefore, $S_{x_0}^\perp$ is the one-dimensional subspace spanned by $\nabla g(x_0)$. Now, $$\nabla f(x_0) = \lambda \nabla g(x_0)\text{ for some } \lambda \in \Bbb{R} \iff \nabla f(x_0) \in S_{x_0}^\perp \iff \nabla f(x_0).v = 0 ...


2

$\cos^2 x+\sin^2 x=1$ is usually a powerful formula. Since \begin{align} (2\lambda x)^2+(2\lambda y)^2 &= 4\lambda^2\\ &=e^{2x}, \end{align} \begin{align} -e^x \sin y + 2\lambda y &= e^x (\pm y-\sin y). \end{align} Since $e^x>0$, $\pm y -\sin y =0$ and $y=0$ is the unique root. Therefore, given function has two extreme point, $(1,0)$ and ...


1

$$ax^{a-1}y^bz^c=\lambda kx^{k-1}$$ $$ax^ay^bz^c=af(x,y,z)=\lambda k x^k$$ Similarly $$bf(x,y,z)=\lambda ky^k$$ And one with $z$ Add all three together $$(a+b+c)f(x,y,z)=\lambda k $$ $$f(x,y,z)=\frac {\lambda k}{a+b+c}$$ However after this part, I think it is quite impossible to find exactly how $\lambda$ depends on $a,b,c,k$. I have tried using ...


1

Just played with it a little and got: $$ \left(\begin{matrix}x^{a}y^{b}z^{c}\\ x^{a}y^{b}z^{c}\\ x^{a}y^{b}z^{c} \end{matrix}\right)=\lambda k\left(\begin{matrix}x^{k}/a\\ y^{k}/b\\ z^{k}/c \end{matrix}\right) $$ Therefore $x^k/a=y^k/b=z^k/c$. By putting it in the constraint, we get $$ 1=x^k+y^k+z^k=(1+b/a+c/a)x^k \\ x=\sqrt[k]{\frac{a}{a+b+c}} $$ and ...


1

The Lagrange multiplier condition tells you that $\nabla f(x_0) = \sum_{i=1}^m \lambda_i\nabla g_i(x_0)$ is normal (orthogonal) to the affine subspace $M$. This means that $\nabla f(x_0)\cdot (y-x_0) = 0$ for any $y\in M$. Thus, since $f$ is convex, for any $y\in M$ we have $$f(y)\ge f(x_0)+\nabla f(x_0)\cdot (y-x_0) = f(x_0).$$ This means $x_0$ is a global ...


1

Hint: substitute the values of t and for (3) use the magnitude to find the distance.


1

There is a second derivative test for constrained extrema. Here's the statement for the case of one constraint $g(x)=c$. Suppose $a$ is a constrained critical point of $f$ on the constraint set $g(x)=c$. Then we have $\nabla f(a)=\lambda\nabla g(a)$ for some scalar $\lambda$. Consider the Hessian matrix $H$ (matrix of second partial derivatives) of ...


1

What's wrong is that $x^2 + y^2 - 1 = 0$ is not the intersection of your two constraints. In fact it has only two points in common with that intersection.


1

Your first equation $-2x=\lambda\cdot 2x$ implies that either $x=0$ or $\lambda=-1$ (or both). Your second equation $2y=\lambda\cdot 2y$ implies that either $y=0$ or $\lambda=1$ (or both). Combining those two, you have a total of four possibilities. Look at each possibility and find the possible points (if any) for that possibility. Then look at the ...


1

Parametrize the curve $g(x) = 0$ with $c(t)$ s.t $c(0) = p$ where $p$ is the local extrema of $f, c'(0) \not = 0$. Then you know that $f(c(t))$ has local min/max when $t = 0$ i.e; $$\frac{d}{dt} f(c(t)) |_{t=0} = \nabla f(p) \cdot c'(0) = 0$$ We also know that $\nabla g(p) \cdot c'(0) = 0$ and so there exists a non-zero scalar $\lambda$ s.t; $$\nabla f(p) ...


1

The level curves of $f$ represent single values of $f$ that increase in a direction parallel to the gradient. This means that, given a level curve that does not represent a local maximum, there is another level curve nearby whose value for $f$ is greater than the first curve. Imagine $g$ as a curve that cuts through a level curve of $f$ at a point $p$. ...


1

We consider $$\Lambda(x,y,z)=x^ay^bz^c-\lambda(x^k+y^k+z^k-1) $$ and by applying $\frac\partial{\partial x}$, $\frac\partial{\partial y}$, $\frac\partial{\partial z}$ are led to the equations $$\begin{align}ax^{a-1}y^bz^c-k\lambda x^{k-1}&=0\\ bx^ay^{b-1}z^c-k\lambda y^{k-1}&=0\\ cx^ay^{b}z^{c-1}-k\lambda z^{k-1}&=0\\ \end{align}$$ If we add ...


1

It only boils down the following statement: For a real symmetric matrix $A$, vector $b$, and $\lambda > -\lambda_\min(A)$ let $$ p(\lambda) = (A + \lambda I)^{-1} b. $$ Then, $\lambda p(\lambda)\to b$ for $\lambda\to\infty$. Proof: Since the inversion of matrices is continuous, we obtain $$ \lambda p(\lambda) = \lambda(A + \lambda I)^{-1} b = ...


1

The critical point (0,b) should not be included as an extrema, since the constraint is $y=x^2$, which is the boundary of the region, whereas (0,b) in inside the region $y=x^2$, so it should not be included. The way we use to determine local max or local min is through Second Derivative Test. Suppose the second derivative of f(x,y) are continuous on a open ...


1

The geometrical picture is the following: We are asked to find the local extrema of the distance from the point $(0,b)$ on the $y$-axis to points on the parabola $y=x^2$. From looking at a figure we can guess the following: If $b\gg1$ there are two local minima high up, and a local maximum at $(0,0)$. If $0<b\ll1$ there is just one local minimum at ...


1

$(2/\sqrt{11})(3,1,-1)$ is the closest point and $(-2/\sqrt{11})(3,1,-1)$ is the most distant point. If you have to use Lagrange multipliiers.... minimize/maximize: $(x-3)^2 + (y-1)^2 + (z+1)^2$ (this is the distance squared from x,y,z to your point.) constrained by: $x^2 + y^2 + z^2 = 4$ $F(x,y,z,\lambda) = (x-3)^2 + (y-1)^2 + (z+1)^2 - \lambda(x^2 + ...


1

The theory of Lagrange multipliers says: Under the given circumstances there is a value $\lambda\in{\mathbb R}$ such that at the conditionally stationary point $x$ the equations $$2Lx-\lambda d=0,\qquad x^Td=k\tag{1}$$ are simultaneously true. Now $(1)$ is a system of $N+1$ equations in $N+1$ unknowns, one of them $\lambda$. If the situation is not ...


1

We can see that $Aa+Bb+Cc+D=0$, so the plane $P:Ax+By+Cz+D=0$ can be written: $$P:A(x-a)+B(y-b)+C(z-c)=0$$ Consider the intersection of $P$ and z-axis. All points on z-axis has the form $(0,0,p)$, then $$A(0-a)+B(0-b)+C(p-c)=0\iff p=\frac{Aa+Bb+Cc}{C}$$ Similarly, two another intersection has coordinate $(\frac{Aa+Bb+Cc}{A},0,0)$ and ...



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