Austin Rangel

2022-10-06

We have just started studying functions of several variables and their derivatives and our professor suggested the following problem as food for thought.

Two squares, both with length l=1 intersect in a rectangle that has an area equal with 1/8 . Find the minimum and maximum distance between the centers of the squares.

Two squares, both with length l=1 intersect in a rectangle that has an area equal with 1/8 . Find the minimum and maximum distance between the centers of the squares.

omeopata25

Beginner2022-10-07Added 5 answers

You need to introduce some coordinates in there to transform that geometry problem into an algebra one.

So let's fix the first square between $(0,0)$ and $(1,1)$, so with center $(1/2,1/2)$. Next, you have $(x,y)$ the coordinates of one edge of the other square, which defines that square up to 4 cases. We will focus on $(x,y)$ being the top right corner.

The center of that second square is then $(x-1/2,y-1/2)$. The distance between the 2 centers is then $d(x,y)=\sqrt{(x-1{)}^{2}+(y-1{)}^{2}}$

So now the problem is to find the extremum of $d(x,y)$ (or better ${d}^{2}$) subject to $xy=1/8$.

With Lagrange multipliers, that's finding the extremum of $\varphi (x,y,\lambda )=(x-1{)}^{2}+(y-1{)}^{2}-\lambda (xy-1/8)$ such that $\frac{\mathrm{\partial}\varphi}{\mathrm{\partial}x}=\frac{\mathrm{\partial}\varphi}{\mathrm{\partial}y}=0=2(x-1)-\lambda y=2(y-1)-\lambda x$

that implies $x=y=\frac{1}{2\sqrt{2}}$ (easy to verify it's a minimum by looking at the 2nd derivative).

As for the maximum, it's obviously not a critical point, so has to be on the border. For instance $x=1$ and $y=1/8$.

So let's fix the first square between $(0,0)$ and $(1,1)$, so with center $(1/2,1/2)$. Next, you have $(x,y)$ the coordinates of one edge of the other square, which defines that square up to 4 cases. We will focus on $(x,y)$ being the top right corner.

The center of that second square is then $(x-1/2,y-1/2)$. The distance between the 2 centers is then $d(x,y)=\sqrt{(x-1{)}^{2}+(y-1{)}^{2}}$

So now the problem is to find the extremum of $d(x,y)$ (or better ${d}^{2}$) subject to $xy=1/8$.

With Lagrange multipliers, that's finding the extremum of $\varphi (x,y,\lambda )=(x-1{)}^{2}+(y-1{)}^{2}-\lambda (xy-1/8)$ such that $\frac{\mathrm{\partial}\varphi}{\mathrm{\partial}x}=\frac{\mathrm{\partial}\varphi}{\mathrm{\partial}y}=0=2(x-1)-\lambda y=2(y-1)-\lambda x$

that implies $x=y=\frac{1}{2\sqrt{2}}$ (easy to verify it's a minimum by looking at the 2nd derivative).

As for the maximum, it's obviously not a critical point, so has to be on the border. For instance $x=1$ and $y=1/8$.

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Now if the function happens to depend on $n$ variables we denote its derivative with respect to the $i$th variable by:

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1) $F(x,y)$ 𝑎𝑛𝑑 $y=f(x)$ so his means that the function $F$ is a function of one variable which is $x$

2) while we were computing 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 we treated $y$ and $x$ as two independent variables although that $y$ changes as $x$ changes but while doing the 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒𝑠 w.r.t $x$ we treated $y$ and $x$ as two independent varaibles and considered $y$ as a constantLet $f:{\mathbb{R}}^{2}\to \mathbb{R}$ be defined as

$f(x,y)=\{\begin{array}{ll}({x}^{2}+{y}^{2})\mathrm{cos}\frac{1}{\sqrt{{x}^{2}+{y}^{2}}},& \text{for}(x,y)\ne (0,0)\\ 0,& \text{for}(x,y)=(0,0)\end{array}$

then check whether its differentiable and also whether its partial derivatives ie ${f}_{x},{f}_{y}$ are continuous at $(0,0)$. I dont know how to check the differentiability of a multivariable function as I am just beginning to learn it. For continuity of partial derivative I just checked for ${f}_{x}$ as function is symmetric in $y$ and $x$. So ${f}_{x}$ turns out to be

${f}_{x}(x,y)=2x\mathrm{cos}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)+\frac{x}{\sqrt{{x}^{2}+{y}^{2}}}\mathrm{sin}\left(\frac{1}{\sqrt{{x}^{2}+{y}^{2}}}\right)$

which is definitely not $0$ as $(x,y)\to (0,0)$. Same can be stated for ${f}_{y}$. But how to proceed with the first part?