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## New answers tagged lagrange-multiplier

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Note that some care must be taken here when applying the Lagrange multiplier method as the cost function is not differentiable at all feasible points. I am assuming that $p_k >0$ for all $k$, and (implicitly) that $m>0$. Note that the feasible set is compact, hence a minimiser and a maximiser exist. Not that it matters here, but the minimum is seen ...

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In the two dimension case, the gradient of $f$ was composed of two equations, one partial derivative wrt $x$ and one wrt $y$. Then $\bigtriangledown L =0$ is in fact a ystem of three equations with three unkonwns $x$, $y$, and $\lambda$. In the $n$ dimensional case, you would have $n$ equations from $\bigtriangledown f (x)=0$ and one equation from $... 0 The minimum is obviously the degenerate case some-lenght = 0. The maximum obviously exists (why?). What can be your solution? 0 Let an isocurve be described by the parametric equation$x=x(t),y=y(t)$. We have $$f(x(t),y(t))=C\implies \frac{df}{dt}=\frac{\partial f}{\partial x}\dot x+\frac{\partial f}{\partial y}\dot y=\nabla f\cdot\vec t=0$$where$\vec t=(\dot x,\dot y)$is a tangent vector. 1 HINT: Directly by symmetry$ x=y=z=a. $0 you have these 4 equations$2x+k(y+z)=02y+k(x+z)=02z+k(x+y)=0xy+yz+zx=3a^2$find$k$from the first:$k=\frac{-2x}{y+z}$replace it in the second and third$2y(y+z)=2x(x+z)2z(y+z)=2x(x+y)y(x+z)+xz=3a^2$rewriting the first and second$(y-x)(x+y+z)=0(z-x)(x+y+z)=0xy+xz+yz=3a^2$from the two first we have$x=y=z$, replacing ... 0 The constraint inequality and constraint just with g(x,y) are similar problems. The difference is that after solution is arrived at by Euler Lagrange, for the inequality we can find a boundary zone when we had only a border line without inequality part$ \lambda(g(x,y) - m) $in operation. 1 Since the volume is fixed, you can express the height as a function of the radius,$y = 4/x^2$. The costs$a$and$b$are also fixed, so you can express the cost of can as a function of$x$. In particular, $$C(x) = 4 \pi b x^2 + \frac{8 \pi b}{x},$$ so $$C'(x) = 8 \pi b \left( x - \frac{1}{x^2} \right).$$ Solving$C'(x) = 0$gives$ x = 1$. The ... 0 To solve for$x$and$y$, you have to plug the result you got after solving for$\lambda$into the constraint$xy=1$, not into the function that you're trying to minimize. In effect, what you've done is wandered off of the$xy=1$curve and instead minimized the function along the line$x=y$. 2 The reason that you get the multiplier condition$\lambda \leq 0$rather than$\lambda \geq 0$is that you have a minimization problem, as you correctly indicate. Intuitively, if the constraint$g(x,y)\leq c$is active at the point$(x^*,y^*)$, then$g(x,y)$is increasing as you move from$(x^*,y^*)$and out of the domain, and$g(x,y)$is decreasing as you ... 1 (1) and (2) give$x(1+8\lambda x^2)=0$and$y(1+18\lambda y^2)=0$, hence$4x^2=9y^2$. Now plug that in the constraint. BTW: nobody cares about$\lambda$'s value. 0 The constraint is an equality constraint. You would need to rewrite it to = 0 and then use a function which worsens the more far off from 0 it becomes. For instance you could use the function $$-(4x^4+9y^4-64)^2$$The minus sign is because "worsening" the function value of a maximisation is always negative. The square makes sure we will always subtract ... 1 Answer: $$y^2 = \frac{\sqrt{64-4x^4}}{3}$$ $$f(x,y) = x^2+y^2 = x^2+\frac{\sqrt{64-4x^4}}{3}$$ Find$\frac{\delta f}{\delta x}$and set it equal to 0 and find$x^2$. Find$y^2$, Add$x^2+y^2 = 4.807$. Using the other method, you get the same value. Good Luck 6$(4x^4+9y^4)(1/4+1/9) \geq (\frac{1}{2}2x^2+\frac{1}{3}3y^2)^2=(x^2+y^2)^2$by Cauchy Schwartz inequality. Now substitute the values to get the answer. When using lagrange multiplier. From your first two equation write$x^2$and$y^2$as a function of$\lambda$.(ignore those x,y=0 solutions for now). And substitute in your third equation. from there you ... 0 The theory of inequality-constrained optimization is very well-developed (particularly for convex problems such as yours). However you can solve your particular problem with the following elementary observation: every extremum of$f$is either on the active set$\partial D$(which you say you know how to solve), or on the interior of the feasible set. Do ... 1 In general, the Lagrange multipliers method gives only a necessary condition for some point to be an extremum of some function$F(x)$subject to the condition$G(x)=c$, where$F,G$are differentiable functions from$\mathbb{R}^n$to$\mathbb{R}$, and$c$some constant. The necessary condition is that at an extremum point, the gradient of$F$needs to be ... 1 Firstly, find the extremae of$f$in the interior of$K$. Then look at the boundary of$K$. In neither of the four cases multipliers are needed as you can solve either equation for$y$and substitute the solution back in$f$. Don't forget to check the edges. Lagrange is needed if it's difficult or impossible to solve the restriction for one variable. 3 This is not an answer since it uses (for illustration purposes) Lagrange multipliers). I am sure that you will love Lagrange multipliers ! So, let me give you a taste for your problem. Considering the function $$F=3x+4y+\lambda (x^2+y^2-14x-6y-6)$$ let us compute the three derivatives $$F'_x=\lambda (2 x-14)+3$$ $$F'_y=\lambda (2 y-6)+4$$ ... 2 Substitute$u=x-7$,$v=y-3.$Then the problem becomes: Find the maximum value of$3u +4v$if$u^2 +v^2 =64.$Now substitute:$u=\frac{k}{3}$and$v=\frac{l}{4}$then our problem becomes: Find the maximum value of$k+l$if$\frac{k^2}{24^2} +\frac{l^2}{32^2} =1$. But we can substitute$k=24\sin \xi, l=32\cos \xi$hence$\$k+l=24\sin \xi +32\cos \xi \leq ...

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