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

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.
Justify your answer.

Prerequisites

Rectangular Hyperbola & it's symmetry
Tangent
Derivative test

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 $(\sqrt{10}, \sqrt{10} )$ .

ISI MStat 2016 Problem 1 graph — Fig-1

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

ISI MStat 2016 Problem 1 Figure 2 — Fig-2

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 $(\sqrt{10}, \sqrt{10} )$ .

ISI MStat 2016 Problem 1 Figure 3 — Fig-3

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 $\sum_{i=1}^{n} x_{i}$ .

Related

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.
Justify your answer.

Prerequisites

Rectangular Hyperbola & it's symmetry
Tangent
Derivative test

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 $(\sqrt{10}, \sqrt{10} )$ .

ISI MStat 2016 Problem 1 graph — Fig-1

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

ISI MStat 2016 Problem 1 Figure 2 — Fig-2

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 $(\sqrt{10}, \sqrt{10} )$ .

ISI MStat 2016 Problem 1 Figure 3 — Fig-3

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 $\sum_{i=1}^{n} x_{i}$ .

Related

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2 comments on “Hyperbola & Tangent | ISI MStat 2016 Problem 1 | PSB Sample”

Sourav Biswas says:

July 30, 2021 at 6:01 pm

Yeah, I went through your process and this was absolutely right. Couldn't it be solved without Calculus? I approached in another way. We have the identity (x+y)2 - (x-2)2 = 4xy. Then (x+y)2>= 4xy. that is |x+y|>= 2 root(xy). in our problem, |x+y|>= 2root(10). That is min|x+y| = 2root(10). Equality will occur iff x=y. then, 2|x|= 2root(10) which gives, |x|=root(10). then also |y|=root(10). Remembering the restriction, x=y, we have the required pairs (x,y) as (root10, root10), (-root10, -root10).
I think this process is quite elegant and Time Saving!
Please let me know if I've made some mistakes. 🙂

Reply
1. Srijit Mukherjee says:
  
  August 2, 2021 at 11:50 pm
  
  Yeah, this one is great and crispy too. 😀 Thanks
  
  Reply

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