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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.

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

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

1. Sourav Biswas says:

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!

1. Srijit Mukherjee says:

Yeah, this one is great and crispy too. ðŸ˜€ Thanks