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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

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 \( \sum_{i=1}^{n} x_{i} \).

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

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 \( \sum_{i=1}^{n} x_{i} \).

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

  1. 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. 🙂

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