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.
(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.
(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
.
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. 🙂
Yeah, this one is great and crispy too. 😀 Thanks