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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 $x \ge \varepsilon, y \ge ... 3 at the points that are closest and the farthest from the plane$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 $$1=(6t)^2+2(3t)^2 + 3(2t)^2 = 66t^2 \to t = \pm 1/\sqrt{66}.$$ ... 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 $$\frac{x^2}{a^2}=\frac{y^2}{2b^2}=\frac{y^2}{2b^2}$$ 2 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$a^2y^2=2b^2x^2$. Substituting this into the constraint: ... 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 ... 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, ... 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 I would like to give you a hint because it is pretty easy. First of all, you need to find the intersection of the two planes (a line). Because you have 2 equations with 3 variables, you can reduce it to 1 equation with 2 variables. Do not forget to impose conditions to make sure two planes do not parallel each other. Secondly, when you have a line in which ... 1 I would like leave a comment. Being a beginner in StackExchange, my reputation is too low to do so. You may solve the equations so that one variable is eliminated and parametrize the solution. As the distance between (0,0,0) and the solution set is given by Pythagoras theorem. you may find the required point by completing square or differentiation. 1 you may be able to do this without the use of lagrange method. let$n_1, n_2$are the unit normal vectors of the two planes. first you can deal with the easier case of$n_1 = \pm n_2.$so wlog we take$n_1 \neq \pm n_2.suppose that the planes $$\frac{a_0}{\sqrt{a_1^2 + a_2^2 + a_3^2}}+ x\cdot n_1\, = 0, \frac{b_0}{\sqrt{b_1^2 + b_2^2 + b_3^2}} + x \cdot ... 1 Since \bar{z} is fixed, the above problem is minimizing a quadratic function f with a single affine constraint. In general, the optimality condition tells us that the gradient of f at the optimal solution is orthogonal to the null space of the affine constraint (or equivalent,lies in the image of the dual of that affine map). Now the gradient of the ... 1 Constrained extrema exist at points \vec r wherever \vec r satisfies the constraint g(\vec r)=\text{ Const. } and...$$\vec\nabla f(\vec r) = \lambda \vec \nabla g(\vec r)$$When \lambda = 0 we have \vec\nabla f(\vec r) =0 , so that \vec r is an extremum of the unconstrained function f Clearly extrema and the unconstrained function will also ... 1 Let x_0 be a particular solution to Ax = b and let M be a matrix whose columns form a basis of the null space of A . Then every solution to Ax = b is equal to x_0 + My for some vector y . So your optimization problem is equivalent to minimizing f (x_0 + My) with respect to y , which is an unconstrained problem. You can ... 1 For Lagrange multipliers,$$\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}$$In this case, ... 1$$x+y=kx^2+4y^2=(k-y)^2+4y^2=5y^2-2ky+k^2=45y^2-2ky+k^2-4=0D=k^2-5(k^2-4)=20-4k^2\geq0k^2\leq5-\sqrt5\leq k\leq\sqrt5x+y=4 \quad vs.\quad x+y=\sqrt5\text{Distance between above two lines }=\frac{4-\sqrt5}{\sqrt2}=2\sqrt2-\frac{\sqrt10}{2}$$1 Any point on the ellipse can be represented as (2\cos t,\sin t) The perpendicular distance$$=\dfrac{|2\cos t+\sin t-4|}{\sqrt2}$$Now 2\cos t+\sin t=\sqrt5\cos\left(t-\arccos\dfrac2{\sqrt5}\right) \implies-\sqrt5\le2\cos t+\sin t\le\sqrt5 \implies-\sqrt5-4\le2\cos t+\sin t-4\le\sqrt5-4 \implies maximum distance =\dfrac{\sqrt5+4}{\sqrt2} ... 1 You can still use Lagrange multipliers, together with the information that \nabla f is perpendicular to the level surface \{f = 1\}. Indeed, it suffices to show that P is parallel to \nabla f(P). To do this we use that P is a solution to the system \begin{cases} \nabla d^2 = \lambda \nabla f \\ f = 1. \end{cases} Notice that the first equation ... 1 First, transform your equation:$$b^2x^2+a^2y^2=a^2b^2a^2y^2=a^2b^2-b^2x^2y^2=b^2-\frac{b^2}{a^2}x^2xy^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=0b^2=3\frac{b^2}{a^2}x^2a^2=3x^2x=\pm\sqrt{\frac{1}{3}}a$$Finally, evaluate xy^2 at x=+\sqrt{\frac{1}{3}}a: ... 1 Note that$$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}$$For the min: xy^2\ge 0 in the given domain and 0 is attained so this is ... 1 You already have the equation of$$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}$$You can then plug in your values that you found for y from solving the second-to-last equation you gave (note that y=0 is a valid solution, as Jack Bauer pointed out, because ... 1 The constraint must be binding, otherwise you could increase the objective function indefinitely. It follows that \lambda \neq 0, and then the argument works. 1 By Cauchy-Schwarz, we have$$4z^2=(1+1)(x^2+9y^2) \ge (x+3y)^2 = (5-3z)^2$$giving$$(z-1)(z-5) \le 0$$which implies$$1 \le z \le 51 \le z^2 \le 25$$The equality holds at (x,y,z)=(1,\frac{1}{3},1) and (x,y,z)=(-5,-\frac{5}{3},5) 1 One thing we need to realize is z^2 is maximized when h(x,y,z)=z is either maximized or minimized. By taking the partial derivatives with respect to x,y,z, you have$$2x\lambda+u=0\implies x=-{u\over2\lambda}18y\lambda+3u=0\implies y=-{u\over6\lambda}-4z\lambda+3u+v=0\implies z={3u+v\over4\lambda}$$Now sub in to x+3y+3z=5 and ... 1 Hint: the linear system$$\eqalign{ \lambda(- 12 y + 2 x + 20) = F_x' &= 0 \cr \lambda(126 y - 12 x + 60) + 9 =F_y'&= 0\cr }$$has a unique solution for 12\lambda^2 + 16\lambda + 1\ne 0 (why?). What happens when 12\lambda^2 + 16\lambda + 1 = 0? 1 I got pretty far..... But the two sets of points i got are equidistant at d = \frac{\sqrt{2}3}{11}=1.2792 I want to use the fact that the distance formula is....$$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 ... 1 Your functionF$is defined on a disc of radius one, which is a compact set. Therefore, the maximums and minimums are either on the boundary of the disc, or strictly inside the disc. Case 1: suppose they are on the boundary. In this case, rewrite your problem in polar coordinates as follows: $$F(r,t)=2r^2\cos^2(t)-3r^2\sin^2(t)-2r\cos(t), \quad ... 1 When exponents get in the way, logarithms come to the rescue. Introduce new variables u=\log x, v=\log y, w=\log z. The problem changes to minimizing$$f(u,v,w)=\frac{1}{p}e^{pu}+\frac{1}{q}e^{qv}+\frac{1}{r}e^{rw}$$subject to the linear constraint u+v+w=\log C. The linearity of constraint is important, because it simplifies the second derivative ... 1 The right values (solving the system by row reduction) are$$ 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. $$1$$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.