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0

I think there is nothing more to say in such a constraint optimisation problem. However, you can check whether the function actually does what it says it does. For instance, to show that they are maximum points, you convert the function to one-variable equation using the constraint, then differentiate once, and substitute. If the values are -ve, then you're ...


1

If any of $x$, $y$, and $z$ is $0$, then two of them are $0$ and the other is not (so as to satisfy $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$). Thus, $\lambda=0$. Hence, all such solutions are $(x,y,z,\lambda)=(\pm a,0,0,0)$, $(x,y,z,\lambda)=(0,\pm b,0,0)$, and $(x,y,z,\lambda)=(0,0,\pm c,0)$. If none of $x$, $y$, and $z$ is $0$, then ...


0

We have: $3(xyz)^2 = 3xy\times yz \times zx = \dfrac{24xyz\times \lambda^3}{a^2b^2c^2}, 3xyz = xyz+xyz+xyz = 2\lambda\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\right)=2\lambda \times 1 = 2\lambda\Rightarrow 2\lambda = 3xyz=\dfrac{24\lambda^3}{a^2b^2c^2}\Rightarrow \lambda^2 = \dfrac{a^2b^2c^2}{12}\Rightarrow \lambda = \dfrac{abc}{2\sqrt{3}}$.


0

The Lagrange function $L(x,\lambda,\mu)$ is affine in $\lambda$ and $\mu$, thus, concave, and the infimum of concave functions is concave. (Equivalently to the supremum of convex is convex.)


0

An interesting challenge arises in working with the Lagrange-multiplier method in that all but the simplest problems (generally those involving linear functions and linear constraint equations) can easily lead to systems of multi-linear or non-linear equations, which often do not have general methods of solution. I find that one must draw on an assortment ...


1

Setting $f_x=\lambda g_x, \;f_y=\lambda g_y, \;f_z=\lambda g_z$ gives $\;\;\displaystyle2x=\lambda\cdot\frac{2x}{a^2}, \;\;2y=\lambda\cdot\frac{2}{b^2},\;\;2 z=\lambda\cdot\frac{2z}{c^2}$. Therefore $\textbf{1)}$ $x=0$ or $\lambda=a^2\;\;\;$$\textbf{2)}$ $y=0$ or $\lambda=b^2\;\;\;$ $\textbf{3)}$ $z=0$ or $\lambda=c^2$ Since ...


0

Applying the indicated "Lagrange equations", we can subtract the second from the first to produce $$ ( \ 2 \lambda \ + \ \frac{1}{2} \ \mu \ ) \ ( \ x \ - \ y \ ) \ = \ 0 \ \ . $$ This makes it a bit more evident that one's intuition about the symmetry of the functions is helpful. As Ted Shifrin already notes, one result from this is that $ \ x \ = \ y \ ...


-1

fx = 50-4x-2y and fy = 48-4y-2x 50-4x-2y=48-4y-2x y=x-1 x+y=24 hours, x+(x-1)=24 SO, x= 25/2 y= 24-25/2 = 23/2 ... f(25/2,23/2)= 200+(25/2)*(50−(25/2))+(23/2)(48−(23/2))−((25/2)+(23/2))^2 =512.5 IS IT CORRECT?


0

If $x$ is the side of the floor, and $y$ the height, and the roof is flat, then indeed we are minimizing $f(x,y)=19x^2+59xy$ subject to the constraint $x^2 y=16000$. This is easily transformed into a $1$ variable problem by substituting $y=\frac{16000}{x^2}$ for $y$ in $19x^2+59xy$. But I gather you want to use Lagrange multipliers. Taking partial ...


2

I know that this question is tagged as a Lagrange multiplier question, but as it turns out, you don't need a Lagrange multiplier because the inequality ($x+y\le 24$) is not tight, so it can be ignored. The derivation is really simple. The total points are given by the equation, $$p=200+x(50-x)+y(48-y)-(x+y)^2.$$ At the maximum, $p$ is extremal with ...


-1

$P(x,y)=200+x(50-x)+y(48-y)-(x+y)^2$ is the function that you want to extremize subject to $x+y=24$. I would solve for $x$ or $y$ in the constraint equation, plug it into $P$, differentiate, set equal to zero and solve.


2

the function you are inserted in is $f(x,y) = 200 +x(50-x) +y(48-y) - (x+y)^2$take the derivative for $x$ and $y$ and set is zero. This will give you the condition for a local extremum. The answer should be $x = 26/3,\ y = 23/3$. Then you still have to explain why that this is a local maximum and not a minimum. And at last you should check the global points. ...


1

A few things: First of all (and most importantly), you call this a Lagrange problem (and your solution does satisfy x + y = 24), but is it really? Seems improbable that the optimal strategy would involve no sleep. I'd look in the interior, $x + y < 24$. Second, I don't get the same numbers you get. I get the same x and y values (using Lagrange, ...


2

Geometric Solution: If $r$ is the shortest distance from $B(0,b)$ to the parabola $P:=\big\{(x,y)\in\mathbb{R}^2\,|\,y=x^2+8\big\}$, then the circle $\Gamma$ centered at $B$ with radius $r$ is tangent to this parabola. Suppose that $A(p,q)$ is a tangential point between $\Gamma$ and $P$. Then, $q=p^2+8$ and the line $\ell$ passing through $A$ tangent to ...


0

Hint The square of the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is effectively given by $$D^2=(x_2-x_1)^2+(y_2-y_1)^2$$ In your case $x_1=0$, $y_1=b$, $x_2=?$, $y_2=x_2^2+8$. So $$D^2=x_2^2+(x_2^2+8-b)^2$$ Minimizing the distance is the same as minimizing the square of the distance. So, you have to consider $$\frac{dD^2}{dx_2}=0$$ and get ...


-3

we have $$d(x)=\sqrt{x^2+(x^2-8-b)^2}$$ and this is only a function in $x$! defining $$f(x)=x^2+(x^2+8-b)^2$$ we get for the first derivative $$f'(x)=2x(17-2b+2x^2)$$


0

Let $g(x,y,z)=x^2+y^2+z^2$. Now, you need to solve the system $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ and $g(x,y,z)=1$. That's the next step in Lagrange multipliers. More precisely, use $\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$ to solve for $x$, $y$, and $z$ in terms of $\lambda$ and then plug those formulae into $g$ and solve for $\lambda$ (Caution, ...


0

I think you may have confused yourself in computing and interpreting your result. First, in solving the set of "Lagrange equations" you found, you can follow the suggestions of Mann or KittyL, or try it this way. We may go ahead and solve each equation for the multiplier $ \ \lambda \ $ thus, $$ \lambda \ = \ \frac{yz}{2x} \ = \ \frac{xz}{4y} \ = \ ...


2

You at least need differentiability for existence of KKT. But if you focus on sufficiency, you can say this: Sufficiency: Suppose your problem is to minimize $f(x)$ over a convex set $X$ and subject to $g_k(x)\leq 0$ for all $k \in \{1, \ldots, K\}$ (call this Probelm P1). Define: $$ L(x, \lambda) = f(x) + \sum_{k=1}^K\lambda_k g_k(x) $$ Assume that ...


1

This is a two-dimensional analogue of something you are probably familiar with: Suppose a differentiable function $f:\mathbb R\rightarrow \mathbb R$ is restricted to an interval $[a,b]$ and you want to find the extrema there. Then, the procedure you follow is: Find the local extrema in $(a,b)$ by checking the values of $x$ in $(a,b)$for which $f'(x)=0$ or ...


1

The theory for inequality constraints, that is $g(x,y) \le 1$, is much more complicated than for equations like $g(x,y) = 1$. As start you could consult the Wikipedia on Karush–Kuhn–Tucker conditions. However, in your simple case $f,g:\mathbb R^2\to\mathbb R$, it boils down to following: You can first try to find a maximizer in the interior, that is, ...


0

A geometric interpretation of this problem is to locate the points of the given triaxial ellipsoid $ \ \frac{x^2}{4} \ + \ \frac{y^2}{5} \ + \ \frac{z^2}{25} \ = \ 1 \ $ , which also lie in the plane $ \ z \ = \ x + y \ $ which lie at the greatest distance from the origin (which we shall find by maximizing the "distance-squared" function $ \ f(x, \ y, \ z) ...


1

As I pointed out in a comment, the problem is quite simply solved using one-variable techniques. However, let us explore your choice of letting $x$ be the ticket price and $y$ the number of tickets sold. Then the number of $5$ dollar increments is $\frac{x-50}{5}$. The resulting number $y$ of tickets sold is given by $$y=1000-10\cdot \frac{x-50}{5}.$$ ...


0

The earning is $RV-FC-VC$, where $RV$ is the revenue. You can find revenue by $X\cdot r$ where $X$ is represented linearly with respect to $r$ using the given conditions. So you will end up with an equation with respect to $r$. I believe you can find the extreme values using first derivative or second derivative test. I hope this can give you a start.


0

This (but the justification) is mechanized in Maple by with(Student[MultivariateCalculus]): LagrangeMultipliers((-u*y+v*x)^2, [-a^2+x^2+y^2, v^2-b^2+u^2], [x, y, u, v], output = detailed); See the long output here exported as a PDF file.


1

I don’t think you even need to take $X,Y$ as you have said. We can define $g:\mathbb{R}^4\rightarrow \mathbb{R}^2$ as $$g(x,y,u,v)=(x^2+y^2-a^2,u^2+v^2-b^2)\equiv(g_1,g_2)$$ For the Lagrange Multipliers Theorem to apply, we need the preimage $g^{-1}(\{0\})$ to be a $2$-dimensional sub-manifold of $\mathbb{R}^2$. It is sufficient to show that ...


0

Your general approach seems to be correct. First we find any critical points inside the region: We define $f(x,y)=\frac{1}{8}xy+200x$. We calculate the partial derivatives $$\frac{\partial f}{\partial x}=\frac{1}{8}y+200$$ $$\frac{\partial f}{\partial y}=\frac{1}{8}x$$ We set both of these equal to $0$ to find our only critical point is $(0,-25)$. But we can ...


1

Let's start with the following question, suppose $$ \Sigma:=\left\{x\in\Bbb R^m \mid f(x)=0\in\Bbb R^q,\quad f(x)\in\mathscr C^1(\Bbb R^m),\quad 1\le q<m \right\} $$ and there is a scalar field, or the so-called "goal function" $\theta(x)\in\Bbb R$ on $\Sigma$. What we are going to do is seek $x_{*}$ such that $$\theta ...


1

The constant does not matter, as only the (total) derivative of $F$ is used in the Lagrange Multipliers method, so additive constants do not affect this. You may include the $c$, or you may not. You could argue that the $F$ found on Wikipedia contains more information about the problem than that in the example, but it proves to be irrelevant to the method. ...


0

Let's assume that you have two independent variables as $x_1$ and $x_2$. To satisfy the constraint equation $g(x_1,x_2)=c\quad$ the change in $g(x_1,x_2)\quad$ must be zero $$dg(x_1,x_2)=\frac{\partial g}{\partial x_1}dx_1+\frac{\partial g}{\partial x_2}dx_2=0$$ $$\longrightarrow dx_2=-\frac{\frac{\partial g}{\partial x_1}}{\frac{\partial g}{\partial ...


4

The "Lagrangian function" $L$ is a purely formal device without any intuitive content, but it condenses the complex geometric data into a simple recipe. The "method" in question relies on the following fact: Given a function $f:{\mathbb R}^n\to{\mathbb R}$ and a surface ("condition") $$S:\quad g(x)=c\ ,$$ the function $f$ cannot assume a conditional local ...


0

I will explain it in 2D. I believe similar argument works in 3D. Let $f(x,y)$ be the objective function, under the constraint $g(x,y)=k$. Suppose $g(x,y)=k$ is a smooth closed curve, and can be parametrized by $(x(t),y(t))$. We are then trying to find the maximum or minimum of $f(x(t),y(t))$. The critical points satisfy $$f_x x'(t)+f_y y'(t)=0$$ which is ...


1

In this case, $q_0$ is all of the "outside" goods, that is, the more it is consumed, the higher the utility will be. This is because the utility function is monotonically increasing in $q_0$, which suggests that the budget constraint should be \begin{equation} q_0 + p_1 q_1 + p_2 q_2 \le Y \end{equation} for budget level of $Y$, and prices of $p_1$ and ...


0

That system of equations does not have that solution for the minimal square of distance, since your suggested point $(\frac 1{\sqrt 2},\frac 1{\sqrt 2})$ does not lie on curve $M_2$. For that point, $x^2+y^2$ is $1$, not $\frac 14$. We do not even need Lagrange multipliers for this. Minimizing the distance between $P_1$ and $P_2$ will do the same as your ...



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