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Since you started off using Langrange multipliers, let’s continue down that path. Using the correct value for $g_y$, we have \begin{align} y^2 &= 2b^2\lambda x \\ 2xy &= 2a^2\lambda y \end{align} which upon eliminating $\lambda$ gives $$a^2y^3 = 2b^2x^2y,$$ so either $y=0$ or $a^2y^2=2b^2x^2$. We can eliminate the former because ...

1

You can still use Lagrange multipliers, together with the information that $\nabla f$ is perpendicular to the level surface $\{f = 1\}$. Indeed, it suffices to show that $P$ is parallel to $\nabla f(P)$. To do this we use that $P$ is a solution to the system \begin{cases} \nabla d^2 = \lambda \nabla f \\ f = 1. \end{cases} Notice that the first equation ...

1

First, transform your equation: $$b^2x^2+a^2y^2=a^2b^2$$ $$a^2y^2=a^2b^2-b^2x^2$$ $$y^2=b^2-\frac{b^2}{a^2}x^2$$ $$xy^2=b^2x-\frac{b^2}{a^2}x^3$$ Next, differentiate and solve to find extrema: $$b^2-3\frac{b^2}{a^2}x^2=0$$ $$b^2=3\frac{b^2}{a^2}x^2$$ $$a^2=3x^2$$ $$x=\pm\sqrt{\frac{1}{3}}a$$ Finally, evaluate $xy^2$ at $x=+\sqrt{\frac{1}{3}}a$: ...

2

just out of curiosity, we know that $x^2/a^2+y^2/b^2=1$, then by AM–GM inequality $$1=\frac{x^2}{a^2}+\frac{y^2}{2b^2}+\frac{y^2}{2b^2}\geq 3\sqrt[3]{\frac{x^2}{a^2}\cdot\frac{y^2}{2b^2}\cdot\frac{y^2}{2b^2}}$$ the equality holds when $$\frac{x^2}{a^2}=\frac{y^2}{2b^2}=\frac{y^2}{2b^2}$$

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You already have the equation of $$g(x,y)=b^2x^2+a^2y^2=a^2b^2$$ You can then write $x$ as a function of $y$: $$x=\pm\frac{1}{b}\sqrt{a^2b^2-a^2y^2}=\pm\frac{1}{ab}\sqrt{a^2-y^2}$$ You can then plug in your values that you found for $y$ from solving the second-to-last equation you gave (note that $y=0$ is a valid solution, as Jack Bauer pointed out, because ...

1

Since $\bar{z}$ is fixed, the above problem is minimizing a quadratic function $f$ with a single affine constraint. In general, the optimality condition tells us that the gradient of $f$ at the optimal solution is orthogonal to the null space of the affine constraint (or equivalent,lies in the image of the dual of that affine map). Now the gradient of the ...

1

For Lagrange multipliers, $$\nabla f(x,y)=\lambda\nabla g(x,y)$$ meaning that $$f_x(x,y)\mathbf{i}+f_y(x,y)\mathbf{j}=\lambda g_x(x,y)\mathbf{i}+\lambda g_y(x,y)\mathbf{j}$$ This gives us two equations: $$f_x(x,y)=\lambda g_x(x,y),\quad f_y(x,y)=\lambda g_y(x,y)$$ We also know from the equation for $g(x,y)$ that $$y=\pm\sqrt{x^2-1}$$ In this case, ...

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Note that $$f(x,y)=xy^2+4\le x(3-x^2)+4=-x^3+3x+4$$ Now for $0\le x\le \sqrt{3}$ you have by differentiating $-x^3+3x+4$ $$-3x^2+3=0 \implies x^2=1 \implies x=1$$ with $(-3x^2+3)'_{x=1}=-6<0$ that it is maximized at $x=1$. So the max is $$x=1, y^2=3-1^2 \implies y=\sqrt{2}$$ For the min: $xy^2\ge 0$ in the given domain and $0$ is attained so this is ...

1

When exponents get in the way, logarithms come to the rescue. Introduce new variables $u=\log x$, $v=\log y$, $w=\log z$. The problem changes to minimizing $$f(u,v,w)=\frac{1}{p}e^{pu}+\frac{1}{q}e^{qv}+\frac{1}{r}e^{rw}$$ subject to the linear constraint $u+v+w=\log C$. The linearity of constraint is important, because it simplifies the second derivative ...

0

$\nabla T = (logy, \frac{x}{y})$ and if you let $g(x,y) = 9x^2 + log^2y$, then $\nabla g = \left(18x, \frac{2log(y)}{y} \right)$. Setting $\nabla T = \lambda \nabla g$, our system of equations is: 1) $log(y) = 18 \lambda x$ 2) $\frac{x}{y} = \frac{2\lambda}{y} log(y)$ 3) $9x^2 + log^2y = 4$ Solve this system.

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This is entirely possible, and happens all the time. In order to guarantee that the subproblem solution obtained at the dual optimum corresponds to the true primal optimum, you need the dual to be differentiable which is equivalent to the primal problem being strictly convex. In the context of linear programming, this is called the "problem of ...

4

When you approach the boundary (that is, when $x \searrow 0$, $y \searrow 0$ or $z \searrow 0$), the function $V$ goes to infinity. Thus you can assume that $V$ admits no maximum on your domain considered If $V$ admits a minimum, it won't be within a $\varepsilon$-range of the boundary. So you can add the conditions $x \ge \varepsilon, y \ge ... 9 The Lagrange multiplier method doesn't tell you what kind of critical point you've found. Usually, we would work around this by saying that the domain is compact and the objective function is smooth, so the minimum is attained at either a critical point or on the boundary. But in this case the domain is not compact, because it's not closed, and your ... 1 Lagrange multipliers $$\nabla f=\lambda\nabla g$$ $$(y^2,2xy)=\lambda(2x,2y)$$$2xy=\lambda2y$gives$x=\lambda$then from$x^2+y^2=8$together withe$y^2=\lambda2x$one have $$x^2-8+2\lambda x=x^2-8+2 x^2=3x^2-8=0$$ so $$x=\pm\sqrt{\frac{8}{3}}$$ and $$y^2=8-x^2=8-\frac{8}{3}=\frac{16}{3}$$ so $$y=\pm\frac{4}{\sqrt3}$$ To find the maximum you have to ... -1 HINT: Use Lagrangian multipliers. $$f(x,y)= \lambda \nabla g(x,y)$$ 0 No, using a matrix will not prevent you from seeing all the points that come out as solutions to the equations, because the equations do not care how you arrange them in order to solve them. And no, using a matrix in general will not produce automatically the minimum, unless there are special reasons for that, as there are in your problem. You ought to be ... 0 First, you don't have to solve a differential equation. The equation you found is valid only at some points. However, with what you did, you're almost done. If$\vec{n}=(u,v,w)$is a unit vector normal to the surface S at a point$P=(x,y,z)$of the surface, you have $$\begin{cases} f(x,y,z)=1 & \text{ as the point is supposed to belong to S}\\ ... 0 Disclaimer: Certainly my sole purpose of this pseudo-answer is to send the OP to (a) right place. Apparently one nice tool to approach these type of problems, is so-called Purkiss principle. Please see here.(answer provided by Henry Cohn) 2 If you're going to substitute like this, be careful to make sure you're not throwing out information when you do. For instance, the initial constraint of x^2 + y^2 + z^2 = 5 implies in particular that y^2 + z^2 \leq 5. If you blindly substitute x^2 = 5-y^2 - z^2 as above, you'll end up trying to maximize yz(5 - y^2 - z^2) with no constraint on y or ... 1$$F(x,y,z,\lambda)=x^{\frac{1}{p}}+y^{\frac{1}{q}}+z^{\frac{1}{r}}+\lambda (xyz-c)$$Thus$$x=[(\frac{pqr}{pq+qr+pr}-1)\times c]^{1/p}y=[(\frac{pqr}{pq+qr+pr}-1)\times c]^{1/q}z=[(\frac{pqr}{pq+qr+pr}-1)\times c]^{1/r}$$May be there are some errors in your deduction. 0 Your initial geometric attempt was good too. The objective function has a gradient equal to (1,2), so the extrema will lie at the intersection between the circle of radius \sqrt{80} and the line y-2x=0. Substituting for y(=2x) in x^2+y^2=80 yields x=\pm4. 1 Let x_0 be a particular solution to Ax = b and let M be a matrix whose columns form a basis of the null space of A . Then every solution to Ax = b is equal to x_0 + My for some vector y . So your optimization problem is equivalent to minimizing f (x_0 + My) with respect to y , which is an unconstrained problem. You can ... 0 I can't be sure, but what I think what you really want to do is use QR decomposition as a method for solving the system of equations that results from 'standard' dual reformulation and the KKT conditions. The Lagrangian is f(x) -\lambda'(Ax-b) The KKT conditions are:$$\nabla f(x) - \lambda'\nabla h(x) = 0 \\ Ax-b = 0$$Since the gradients form a ... 2 The first order conditions require differentiability so you can't use those. What you really need is the generalized KKT conditions, that deal with sub-differentiability. Basically, rather than the gradient of the Lagrangian to be zero at x^*, you need the zero vector to be a subgradient at x^*. This is sufficient since your problem is still concave, ... 0 Consider the problem to minimize f(x) = x^2 such that x=3. There is only one solution, namely x=3. But \nabla f(x) \neq 0 there. 1 I got pretty far..... But the two sets of points i got are equidistant at d = \frac{\sqrt{2}3}{11}=1.2792 I want to use the fact that the distance formula is....$$d^2 = (x-u)^2+(y-v)^2+(z-w)^2 $$So i want to maximise and minimise u,v,w. And i want to use two constraints$$f(x,y,z,u,v,w) = (x-u)^2+(y-v)^2+(z-w)^2 h(x....w) = u+v+w-10 = 0 \quad ... 3 at the points that are closest and the farthest from the plane$x+y+z = 10$should have the normal$(2x, 4y, 6z)$of the surface$x^2 + 2y^2 + 3z^2 = 1$be parallel to the normal$(1,1,1)$of the plane. therefore we can take$x = 6t, y = 3t, z= 2t.$making this point on the ellipsoid requires $$1=(6t)^2+2(3t)^2 + 3(2t)^2 = 66t^2 \to t = \pm 1/\sqrt{66}.$$ ... 1 The right values (solving the system by row reduction) are $$x=\frac{16}{15},\ y=\frac13\, z=-\frac{11}{15},\ \lambda_1=-\frac{52}{75},\ \lambda_2=-\frac{18}{25}.$$ Plugging the values you can check that you made a mistake in your second equation for the lambdas. It should have been $$3\lambda_1-14\lambda_2-8=0.$$ 2 By inequality$2x^2+2y^2\geq (x+y)^2$we indeed have$-\sqrt2 \leq x+y \leq \sqrt{2}$and to maximize/minimize$z$we need to minimize/maximize$x+y$so your solution until$x=y=\pm{1\over\sqrt2}$is correct. However how do you get$z$looks unreasonable as$x+y+z=1$but your solution, for example${1\over\sqrt2}+{1\over\sqrt2}+{1\over\sqrt8}$does not ... 2 No, this is not correct. It looks like you assumed that the first function described in the problem was the objective function (that which is to be optimized) and the second function is the constraint function. But if you read the problem, you see that both$x^2 + y^2 = 1$and$x+y+z=1$are constraints. So you need a Lagrange multiplier approach with two ... 1 This topic has probably long been forgotten, but in the interest of the potential future readers let me just point out the fact that the OP's approach did not fail at all. It seemed to fail for the trivial reason that the derivatives were calculated incorrectly. Using the OP's notation (and the convention of regarding$z$and$z^*$as independent variables), ... -1 during the step Replacing λ in F'y, you divided the denominator by 4 because of 4λ. Can you actually do this? It seems like it would change the value. 0 Consider $$F=x^{7/10} y^{3/10}+\lambda (2 x+4 y-3)$$ compute the derivatives and set them equal to zero $$F'_x=2 \lambda +\frac{7 y^{3/10}}{10 x^{3/10}}=0$$ $$F'_y=4 \lambda +\frac{3 x^{7/10}}{10 y^{7/10}}=0$$ $$F'_\lambda=2 x+4 y-3=0$$ From$F'_x=0$, you can extract $$\lambda=-\frac{7 y^{3/10}}{20 x^{3/10}}$$ Replacing$\lambda$in$F'_y\$, you get ...

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