Minima
For real numbers x and y , find the minimum value of 2 x 2 + 2 x y + y 2 − 2 x + 2 y + 4 .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
For a perfect solution,you must show that there is
minima
at
(
x
,
y
)
=
(
2
,
−
3
)
and not
maxima
.
Still, i like your approach,so upvoted.
Your method is very much limited.
Can you solve
This
? from your method
There are flaws in your solution. As mentioned by akhil , only differentiating won't assure that those critical points are minima. It might be a saddle point even. Here is how to justify your r solution.
f x x = 4 = A , f x y = 2 = B , f y y = 2 = C . where f(x,y) denotes the function & f x x signifies partially differntiating f(x,y) twice with respect to x.
We know by axioms that minima occurs if ,
A C − B 2 > 0 , A > 0 and it is satisfied so we do have a minima here .
Solving f x = f y = 0 as you did , now you can add confidence to your statement that indeed its the minimum and it occurs at (2,-3)
f ( x , y ) = ( x + y + 1 ) 2 + ( x − 2 ) 2 − 1 ≥ − 1 , w i t h e q u a l i t y a t ( 2 , − 3 )
Use Partial Differentiation.
Differentiating with respect to x ,
4 x + 2 y − 2 = 0
Differentiating with respect to y ,
2 x + 2 y + 2 = 0
Solving both equations, we get: x = 2 , y = − 3
Hence the minimum value of f ( x , y ) = 8 − 1 2 + 9 − 4 − 6 + 4 = − 1