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7

HINT: Use $$\begin{cases} f_x = \lambda g_x \\ f_y = \lambda g_y \\ xy = 3\end{cases} $$ Note that $f_x = 2x, f_y = 2y, g_x = y, g_y = x$.


2

Here's another approach for fun. We can use the AM-GM mean: $$\sqrt{ab}\leq \frac{a+b}{2},\quad \text{for all}\ a,b\geq 0$$ where equality holds if and only if $a=b$. Use $a=x^2,\ b=y^2$ to get: $$\sqrt{x^2y^2}\leq \frac{x^2+y^2}{2},$$ and our minimum will be where $x=y$. This implies $x=y=\pm\sqrt{3}$ and thus: $$2\sqrt{(xy)^2} = 2|xy| = 6\leq x^2+y^2,$$ ...


2

$ A = x^{4} + y^{4} - 4b^{2}xy + 2b^4$ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2x^{2}b^2 + 2y^2b^2 - 4b^{2}xy $ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2b^2(x^2 + y^2 -2xy)$ $A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2b^2(x - y)^2$ Which is positive as a sum of squares


2

Hints: let $f(x,y) = x^4+y^4-4b^2xy$. Show that the critical points (gradient equals zero) of $f$ are $(b,b)$ and $(-b,-b)$, show that they are local minima, and note that $f$ at these points takes the value $-2b^4$.


2

you must solve the system $$2x+\lambda(1+2x)=0$$ $$2y+\lambda(1+2y)=0$$ $$x+y+x^2+y^2=12$$ easy you will get $$x=y=-3,\lambda=-\frac{6}{5}$$ or $$x=y=2,\lambda=-\frac{4}{5}$$


2

Your solution is correct, except the last part. You should get four minimizing points. (You mistakenly assumed that $y$ and $z$ must be both positive or both negative.) However, there is a solution without using Lagrange multipliers. Note by AM-GM that $$x^2+y^2+z^2=x^2+2\left(\frac{y^2}{2}\right)+4\left(\frac{z^2}{4}\right)\geq ...


2

If you realize that the distance between the lines $x+y=k$ and $x+y=2$ is just $\frac{|k-2|}{\sqrt{2}}$ the problem boils down to computing the minimum and maximum of $x+y$ over $x^2+2y^2=1$. Lagrange's method gives $x=2y$, hence $(x,y)=\pm\frac{1}{\sqrt{6}}(2,1)$ and $k=\pm\sqrt{\frac{3}{2}}$.


2

Your extremal points are correct, but you made rather more work for yourself than you needed to. The Lagrange-multiplier method does tell us about the extremal points on the constraint circle. So we find a minimum value $ \ -75 \ $ at $ \ (-3, \ 4) \ $ and a maximum of $ \ 125 \ $ at $ \ (3, \ -4) \ $ . You are also correct in saying that we are not ...


2

$$f = xy^2$$ with $$4=x^2+4y^2$$ becomes $$f = xy^2 = x\left(1-\frac{1}{4}x^2\right) = x - 0.25x^3$$ maximum via derivative $$\frac{d}{dx}f = 0 = 1 - 0.75x^2$$ $$x_{1,2}=\pm\sqrt{\frac{4}{3}}$$ $$\frac{d^2}{dx^2}f(x_1) < 0 $$ $$\frac{d^2}{dx^2}f(x_2) > 0 $$ and with $4=x^2+4y^2$ again: $$y_{1,2} = \pm\sqrt{\frac{2}{3}}$$ I think that's ...


2

You can suppose $x,y>0$ otherwise $xy^2\leq 0$. Then you can use that $\frac{x_1+x_2+x_3}{3}\geq \sqrt[3]{x_1x_2x_3}$ and the equality holds iff $x_1=x_2=x_3$. Note that $(x^2+2y^2+2y^2)^3\geq 27x^2y^4$. So $64\geq 27(xy^2)^2$ the eqaulity holds iff $x=\sqrt{2}y$ or $x=-\sqrt{2}y$. Therefore $xy^2\leq \frac{8}{3\sqrt{3}}$.


2

Write $y(t)=\cos\big(x(t)\big)$ and $z(t)=\sin\big(x(t)\big)$. Therefore, we may take $x(0)=0$ and $x(1)=\left(2n+\frac12\right)\pi$, for some integer $n$. The integral becomes $\int_0^1\,\sqrt{1+\big(x'(t)\big)^2}\,\text{d}t$, which calculates the length of the curve $\big(t,x(t)\big)$ from $t=0$ to $t=1$. The critical curves are therefore straight ...


2

I'm not entirely sure what you're getting at, but I think this maybe helpful to you: The intuition is that an extreme point of $f(x)$ under the condition $g(x) = 0$ must satisfy $g(x) = 0$ and $\nabla_\nu f(x) = 0$ for any direction $\nu$ tangential to the candidate set $\{g(x) = 0\}$. If this were not the case, we could go a small, positive distance along ...


1

The simple solution (without KKT): it is easy to see that the problem is actually decoupled, i.e. the condition $0<x_k\le q$ affects only one term $\log_2x_k$ in the objective function. It makes it possible to minimize over $x_k$ each term independently, i.e. $$ \min\sum_{k=1}^n(-\log_2 x_k)=\sum_{k=1}^n\min_{0<x_k\le q}(-\log_2x_k)= ...


1

Let $ g(x,y)= x^2+y^2-25$. The Lagrange multiplier and required points can in other words be evaluated by: $$ \lambda = \dfrac{f_x}{g_x}=\dfrac{f_y}{g_y} $$ Second order derivatives ( 6 in total) can be used to further establish min/max as with the single independent variable.


1

Your set compact set $K$ is stratified: It is a $3$-simplex consisting of an open interior, $4$ relatively open two-dimensional facets, $6$ relatively open edges, and $4$ vertices. In order to find the extrema of $h$ on $K$ we have to look at the $1+4+6+4=15$ faces in turn, in order to set up a "candidate list". As $$\nabla ...


1

Let the cost of 1 square metre of plywood be 1 (the actual amount is irrelevant, we just need to set a base unit). So the cost of 1 square metre of glass is 2. Now let the dimensions of the box be $x$, $y$, and $z$ metres, with the glass side being one of the two $x \times y$ sides. Then the cost of the box is $C(x, y, z) = 3xy + 2xz + 2yz$ (because you ...


1

For (a), $1$ is a regular value of $f:\mathbb{R}^3\to\mathbb{R}$ defined by $\left(x_1,x_2,x_3\right)\mapsto x_2x_3+x_3x_1+x_1x_2$. That is, at every point $p$ in $f^{-1}(1)$, $f$ is a submersion at $p$. In other words, $\text{d}_pf:\text{T}_p\mathbb{R}^3\to\text{T}_1\mathbb{R}$ is surjective. This can be easily seen since the matrix representing ...


1

If $f(x,y) = x + y$ then $\nabla f (x,y) = (1,1)$ orthogonal to the line $x + y = 2$. Considering $\varphi (x,y) = x^2 + 2y^2$ we have that $\nabla \varphi (x,y) = (2x , 4x)$. Notice that $1$ is a regular point of $\varphi$ and $M = \varphi^{-1} (1)$ is an ellipse, therefore compact. The restriction $f|_M$ has then at least two critical points (Weiestrass ...


1

Another possibility. First, $$\tag{$*$} xy^2z^4 = 32 \Longrightarrow y^2 = \frac{32}{xz^4}. $$ Notice that we can arbitrarily divide by $x,y$, or $z$ since any point with a coordinate equal to $0$ do not belong to your surface. Now, consider the squared norm of a generic point $(x,y,z)$ and use $(*)$: $$ x^2+y^2+z^2 = x^2 + \frac{32}{xz^4} + z^2 =: N(x,z). ...


1

There was nothing "wrong" with your use of the Lagrange multipliers, but the method often leaves us with a set of equations that need to be "handled the right way" in order to be fully useful. (There is no general method for finding this "right way", since there are many ways that the system of equations can present us with an algebraic tangle. I am also ...


1

Hint Highest (lowest) point is when $z$ is maximal (minimal). Since you are on the paraboloid, $z$ is fixed as a function of $x,y$. Plug this back into your constraint to get to optimize $f(x,y)$ subject to $g(x,y) = 12$.


1

The method of Lagrange's multipliers is a theorem with a few assumptions. Check it out on your favorite advanced calculus book. But, what is even more important, this method works for constrained critical points and should not be seen as an alternative to solving $\nabla f=0$.


1

There is also a geometric solution: the graph of $xy=3$ is a rectangular hyperbola, so we'll have to find the closest points of the hyperbola from the origin.



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