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# Hyperbola & Tangent | ISI MStat 2016 Problem 1 | PSB Sample

This is a beautiful problem from ISI MStat 2016 (sample ) PSB Problem 1. This is based on finding the minimum value of a function subjected to the restriction .

This is a beautiful sample problem from ISI MStat 2016 PSB Problem 1. This is based on finding the minimum value of a function subjected to the restriction.

## ISI MStat 2016 Problem 1

Let $x, y$ be real numbers such that $x y=10$ . Find the minimum value of $|x+y|$ and all also find all the points $(x, y)$ where this minimum value is achieved.

## Solution

(a) Using graph

The equation $xy=10={(\sqrt{10})}^2$ represents the equation of rectangular hyperbola with foci are $(- \sqrt{10},- \sqrt{10})$ and $(\sqrt{10}, \sqrt{10} )$ .

Now , $|x+y|=c \Rightarrow$ $x+y=\pm c$ , which looks somewhat like this ,

we have to find the the minimum value of $|x + y|$ subject to the restriction that $xy=10$ . If we move $|x+y|=c$ along $xy =10$ by varying c , then local minimum can occur at the points where the level curve $|x+y|=c$ touch $xy =10$ . Now as both the rectangular hyperbola and |x+y|=c are symmetric about $x+y=0$ for $c \ne 0$ and $x=y$ , the level curve will touch $xy =10$ when x+y=c and x+y=-c both are tangent to the curve $xy =10$ . And tangents occurs at the foci of $xy =10$ i.e at $(- \sqrt{10},- \sqrt{10})$ and $(\sqrt{10}, \sqrt{10} )$ .

Hence , the minimum value of $|x + y|$ are $|\sqrt{10} +\sqrt {10} |$ and $|-\sqrt{10}- \sqrt{10}|$ both gives the same value $2 \sqrt{10}$ .

Therefore , the minimum value of $|x + y|$ is $2 \sqrt{10}$ and it attains it’s minimum at $(\sqrt{10} , \sqrt{10} )$ and $( – \sqrt{10} , -\sqrt{10})$ .

(b) Using Derivative test

$|x+y|= |x+ \frac{10}{x} |$ as we are given that $xy=10$

Let, $f(x) =|x+ \frac{10}{x}|$ then we have to find the minimum value of f(x)

Now ,as the function is not defined at x=0 and also x=0 can’t give the minimum value of |x+y| due to the condition that xy=10. So, we will study f(x) for two cases when x>0 and when x<0 .

$f(x) = \begin{cases} x+ \frac{10}{x} & ,x > 0 \\ -(x+ \frac{10}{x} ) & ,x < 0 \end{cases}$

$f'(x)=\begin{cases} 1- \frac{10}{x^2} & ,x > 0 \\ -(1- \frac{10}{x^2} ) & ,x < 0 \end{cases}$

$f'(x)=0 \Rightarrow$ $x= \pm \sqrt{10}$

$f”(x)= \begin{cases} \frac{20}{x^3} & ,x > 0 \\ -\frac{20}{x^3} & ,x < 0 \end{cases}$

So,$f”(x) >0$ for $x= \pm \sqrt{10}$

Hence f(x) attains it’s minimum value at $(\sqrt{10} , \sqrt{10} )$ and $– \sqrt{10} , -\sqrt{10})$ and minimum value is $2 \sqrt{10}$.

## Challenge Problem

Let $x_1, x_2 ,…, x_n$ be be positive real numbers such that $\prod_{i=1}^{n} x_{i} = 10$ . Find the minimum value of $\sum_{i=1}^{n} x_{i}$.

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