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4

METHODOLOGY $1$: As suggested in the comment from @EwanDelanoy, we can use the Cauchy-Schwartz Inequality when $x_i>0$ for all $i=1,\cdots,n$ by writing $$\begin{align} n^2&=\left(\sum_{i=1}^n (1)\right)^2\\\\ &=\left(\sum_{i=1}^n \sqrt{x_i}\,\frac{1}{\sqrt{x_i}}\right)^2\\\\ &\le \left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n ...


4

The consumer is solving $$\max x_1^{1/3}x_2^{1/2}$$ subject to $$2 x_1 + 5 x_2 \leq 40.$$ It should be clear that the constraint must bind ($2x_1 +5x_2=40$), since the objective is increasing in both $x_1$ and $x_2$. Using the method of lagrange multipliers, any optimal consumption plan must satisfy the following FOCs: $$1/3 x_1^{-2/3} x_2^{1/2}=2 ...


3

I can provide a solution without using a Lagrange Multiplier. Let us call the closest point on the parabola to the line $\left(x_0,x_0^2\right)$. Then $2x_0=1$, since the tangent line to the parabola at that point must be parallel to the line $y-x+2=0$. (If it is unclear why, imagine rotating and translating the graph so that the line $x-y+2=0$ becomes the ...


3

The relationship between the two formulations is that the partial derivatives of the second formulation give the vector components of the first, and the constraint $g(\textbf x)=0$. To use the approach given by the second formulation, one finds the partial derivatives of $L$ with respect to $x_1, x_2, \dots$ and $\lambda$: $$\frac{\partial L}{\partial ...


3

If $z=0$ you have: $$ \begin{aligned} 2x&=-\lambda y\\ 2y&=-\lambda x\\ xy&=-1 \end{aligned} $$ and the first two of these imply that $\lambda^2=4$ and that $x=y$ or $x=-y$ and the first of these is incompatible with $x$ and $y$ being real and satisfying the third equation. So you are left with $x=-y$ and $xy=-1$... If $z \ne 0$ we have ...


2

The thing is like in the following picture. of wikipedia. For $f=d$ you increment $d$ until you touch $g=c$. In the moment of contact you take a minimum. If you go on, just before $f=d$ leaves the contact, you take the maximum. Thinking it well it is like parking! Really, the idea is so productive that is the base of the Morse's theory.


2

Your setup is correct, in as much as you do indeed want to solve the system of three equations $$\nabla f = \lambda \nabla g$$ $$g = 0.$$ However you have to be extremely careful when doing so, as generally you will have many local extrema and saddle points in addition to the global extrema you search for. It is very easy to accidentally drop possible ...


2

There are two ways of interpreting this question. The first is "what is the maximum possible product for a (finite) sequence of natural numbers whose sum is 20". The second is "Fix the natural number $n$. What is the maximum possible product for a sequence of $n$ natural numbers whose sum is $n$". I assumed the first interpretation was correct. In which ...


2

At a stationary value of the objective $f(x,y,z)$, subject to constraint $g(x,y,z)=0$, the level surface of the objective is kissing the constraint surface. This means the normals to the two surfaces are parallel which is expressed by there existing a real scalar $\lambda$ such that $\nabla f=\lambda \nabla g$. The gradient $\nabla h(\bf{x})$ is a vector ...


2

The extremal temperatures are taken either in the interior $D$ or on the boundary $\partial D$ of the unit disk. Writing $$T(x,y)=2x^2+\left(y-{1\over2}\right)^2-{1\over4}$$ we see that the graph of $T$ is a paraboloid with its apex at $P_1=\bigl(0,{1\over2}\bigr)\in D$, and there are no other stationary points of $T$ in $D$. Take note of ...


2

From \begin{align*} x &= \lambda(x-3)\\ y &= \lambda(y-2) \\ z &= \lambda(z-1) \end{align*} we see that $z \neq 1$ and $\lambda = z/(z-1)$ whence $y(z-1) = z(y-2)$ which gives $y = 2z$. Likewise $x(z-1) = z(x-3)$ or $x = 3z$. Plugging these into the constraint equation $$(x-3)^2 + (y-2)^2 + (z-1)^2 = 1$$ gives $14(z-1)^2 = 1$ or $z = 1 \pm ...


1

You wish to maximize $A = f(x,y) = {1 \over 2} (2 x) (2 y) = 2 x y$ subject to the constraint $g(x,y) = x^2 + 4 y^2 - 1 = 0$. Form the Lagrangian: ${\cal L}(x,y,\lambda) = f(x, y) - \lambda g(x,y) = 2 x y - \lambda (x^2 + 4 y^2 - 1)$. Take the three derivatives and set them to zero: ${\partial {\cal L}(x,y,\lambda) \over \partial x} = 2 y - 2 \lambda x ...


1

I can provide a solution without using Lagrange multipliers. $2x^2+y^2-y=2x^2+(y-\frac{1}{2})^2-\frac{1}{4}$ The coldest point is $\left(0,\frac{1}{2}\right)$, where the temperature is $-\frac{1}{4} $ The hottest point must be somewhere along the circle $x^2+y^2=1$, since the further away the a point is from $\left(0,\frac{1}{2}\right)$, the hotter it is. ...


1

This is a bad exercise, since the answer is much easier to get by noting that we're looking for the points closest to and farthest from the origin on the sphere of radius $1$ around the point $u=(3,2,1)$, which are $u\pm\frac u{|u|}$. If you really want to do it the hard way, you can solve, say, your third equation for $\lambda$, substitute that into the ...


1

We are asked to maximize $U(x_1,x_2)$ subject to $2x_1+5x_2=40$. We can rearrange the constraint to get $x_1=\frac 12(40-5x_2)$. We might as well maximize $U^3(x_1,x_2)=x_1x_2=\frac 12(40-5x_2)x_2$. Now take the derivative, set to zero, and you will have the value for $x_2$.


1

Write $$T (x,y) = 2 x^2 + y^2 - y = 2 x^2 + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} \geq -\frac{1}{4}$$ Thus, the coldest point is $(0,\frac{1}{2})$, where the temperature is $-\frac{1}{4}$. The hottest point must be on the circle $x^2 + y^2 = 1$, which is the domain of interest's boundary. Introduce the Lagrangian $$\mathcal{L} (x,y,\lambda) = T ...


1

In this case, simple geometric considerations tell you that you have a minimum: the objective function is the square of the distance from the origin and the constraint is the equation of a plane. More generally, one can examine the bordered Hessian to determine the nature of the stationary points found via Lagrange multipliers. In this case, $Hf = 2I_3$ ...


1

Well you can always compare it to other values. For example, take $x=1, y=1, z=10$, $x+y+z=12$, and $f(x)=1+1+100=102>48$ So 48 has to be a minimum.


1

The Lagrangian is $$L(x,\lambda) = \tfrac{1}{2} x^T Q x + \lambda^T ( A x - b)$$ The optimality conditions are $$\begin{bmatrix} Q & A^T \\ A & 0 \end{bmatrix} \begin{bmatrix} x \\ \lambda \end{bmatrix} = \begin{bmatrix} 0 \\ b \end{bmatrix}$$ This is a symmetric indefinite linear system and can be solved for the combined value of $(x,\lambda)$. But ...


1

It is correct. Just a quick reminder: Note that, in general, for inequality constraint $g(x) \leq 0$, we would consider $g(x)\mu = 0$. Hence, we should have $$\lambda(x-y+2z-12)=0\\ \mu(x+2y+3z-8)=0 .\\$$ But that is the same as what you have written down.


1

You asked where exactly you have made a mistake: Multiplying equations 1,2,3 by $x,y,z$ we obtain that $2x^2=4y^2=6z^2$. This is correct. (And, since you know what the solution should be, you can check for yourself, that the solution fulfills this equation.) From here I found the corresponding multiples that $x^2$ and $y^2$ are in terms of $z$ and ...


1

By CS inequality: $$ x+y=(x,y)\cdot (1,1)\le \sqrt{x^2+y^2}\sqrt{2} $$ Since $x+y=3$: $$ 3\le \sqrt{2}\sqrt{x^2+y^2} $$ Now, squaring both sided yields $$ 9\le 2(x^2+y^2) $$ In other words $$ x^2+y^2\ge \frac{9}{2} $$ This lower bound is attained when $x=y=3/2$.


1

The map $$ \left(x,y\right)\mapsto \left(\frac{x+y}{\sqrt{2}},\frac{x-y}{\sqrt{2}}\right) $$ is an isometry of $\mathbb{R}^2$, hence your problem is equivalent to finding the extreme values of $\frac{u^2-v^2}{2}$ over $1\leq u^2+v^2\leq 4$. They clearly are $\color{red}{\pm 2}$, and they clearly occur at points of the outer boundary, also because $f(x,y)=xy$ ...


1

I have not been able to find the original question and I give you here my approach hoping it will help you. We want to maiximize $xyz$ subject to the constraint $x^2+2 y^2+3 z^2=a$ with ($x\geq 0$) , ($y\geq 0$) , ($z \geq 0$). So, let us consider $$F=x y z +\lambda \left(x^2+2 y^2+3 z^2-a\right)$$ Computing derivatives $$F'_x=y z+2 \lambda x=0\tag 1$$ ...


1

An appropriate Lagrangian is $$L(x,y,z,\lambda,\mu)=x^2+y^2+z^2+\lambda(x^2+y^2+2z^2-5)+\mu(x+2y+z-5).$$ The nontrivial partial derivatives are: $$L_x = 2x+2\lambda x+\mu \\ L_y=2y+2\lambda y+2\mu \\ L_z=2z+4z\lambda+\mu.$$ If $\lambda,\mu$ were given, these conditions would allow you to find $x,y,z$: indeed ...


1

The inequality also follows directly from the fact that the arithmetic mean is greater than the harmonic mean $\frac{x_1+\ldots+x_n}{n}\geq \frac{n}{\frac{1}{x_1}+\ldots+\frac{1}{x_n}}$.


1

The Lagrangian equations are $$\begin{cases}\begin{align}2x&=-\lambda y\\2y&=-\lambda x\\2z&=2\lambda z\\z^2&=xy+1\end{align}\end{cases}$$ A systematic approach is by noting that the first three form a parameteric linear system with determinant ...


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

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) ...



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