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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.

Let be real numbers such that . Find the minimum value of and all also find all the points where this minimum value is achieved.

Justify your answer.

- Rectangular Hyperbola & it's symmetry
- Tangent
- Derivative test

(a) Using graph

The equation represents the equation of rectangular hyperbola with foci are and .

Now , , which looks somewhat like this ,

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

Hence , the minimum value of are and both gives the same value .

Therefore , the minimum value of is and it attains it's minimum at and .

(b) Using Derivative test

as we are given that

Let, 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 .

So, for

Hence f(x) attains it's minimum value at and and minimum value is .

Let be be positive real numbers such that . Find the minimum value of .

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.

Let be real numbers such that . Find the minimum value of and all also find all the points where this minimum value is achieved.

Justify your answer.

- Rectangular Hyperbola & it's symmetry
- Tangent
- Derivative test

(a) Using graph

The equation represents the equation of rectangular hyperbola with foci are and .

Now , , which looks somewhat like this ,

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

Hence , the minimum value of are and both gives the same value .

Therefore , the minimum value of is and it attains it's minimum at and .

(b) Using Derivative test

as we are given that

Let, 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 .

So, for

Hence f(x) attains it's minimum value at and and minimum value is .

Let be be positive real numbers such that . Find the minimum value of .

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Why Cheenta?

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. ðŸ™‚

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