# Tag Info

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 ...

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 $$k+l=24\sin \xi +32\cos \xi \leq ... 2 The reason that you get the multiplier condition \lambda \leq 0 rather than \lambda \geq 0 is that you have a minimization problem, as you correctly indicate. Intuitively, if the constraint g(x,y)\leq c is active at the point (x^*,y^*), then g(x,y) is increasing as you move from (x^*,y^*) and out of the domain, and g(x,y) is decreasing as you ... 1 Since the volume is fixed, you can express the height as a function of the radius, y = 4/x^2. The costs a and b are also fixed, so you can express the cost of can as a function of x. In particular,$$ 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). $$Solving C'(x) = 0 gives  x = 1. The ... 1 Firstly, find the extremae of f in the interior of K. Then look at the boundary of K. In neither of the four cases multipliers are needed as you can solve either equation for y and substitute the solution back in f. Don't forget to check the edges. Lagrange is needed if it's difficult or impossible to solve the restriction for one variable. 1 In general, the Lagrange multipliers method gives only a necessary condition for some point to be an extremum of some function F(x) subject to the condition G(x)=c, where F,G are differentiable functions from \mathbb{R}^n to \mathbb{R}, and c some constant. The necessary condition is that at an extremum point, the gradient of F needs to be ... 1 HINT: Directly by symmetry  x=y=z=a.  1 Answer:$$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

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.

Only top voted, non community-wiki answers of a minimum length are eligible