Minimize this
For real numbers x , y and z , find the minimum value of 2 x 2 + 5 y 2 + 2 z 2 + 4 x y − 4 y z − 2 z − 2 x .
The answer is -1.
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4 solutions
Nice application of calculus. Congratulations.
The expression can be put as under:-
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Square can not contribute to any negative values.
But giving any value to y will ultimately give a positive
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∴
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This is a parabola opening up, with vertx as
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Yes ! I was able do it without calculus ! Nice solution ! Upvoted.
Nice solution! Without calculus! :)
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I still prefer calculus because it is general. There is a reason why mathematicians discovered it.
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@Steven Zheng – Yeah! I prefer calculus as well. And I was wondering why is it kept under algebra and not calculus.
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@Pranjal Jain – This problem can be solved using both calculus and algebra, but I find the algebra solution more elegant. Hint:
The expression can rewritten as ( x + 2 y − z ) 2 + ( x + z − 1 ) 2 + y 2 − 1
See the elegance? Although this can sometimes be problematic (this happens when we get an inconsistent system)
I also did the same. but pranjal can u explain why we partially differentiate to get max.
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Since x,y,z are independent of each other, we can partially differentiate the given expression. Had there been any restriction, I would have done this and checked if it satisfies restriction. Else, I must have used some other method like Langrange's multipliers or something else. It can also be explained by using some concepts of topology, I suppose.
What I am trying to do is calculating x such that "surface" of expression has "slope" 0. Similarly y and z. And then finding a point(s) in 3 dimensions where it gets "slope" 0 with change in x, y or z.
"Slope" used here is not the one used in 2 dimensions.
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you are explaining me the reason which I explained u in van...great
If you want to solve by only using algebra...
this can also be written as,
S = 2 ( x + y ) 2 + 2 ( y − z ) 2 + y 2 − 2 ( z + x ) l e t ( x + y ) = a , ( y − z ) = b , ( z + x ) = a − b w h e r e a , b , y a r e m u t u a l l y i n d e p e n d e n t v a r i a b l e s S = 2 a 2 + 2 b 2 − 2 a + 2 b + y 2 = 2 ⎩ ⎨ ⎧ ( a − 2 1 ) 2 + ( b + 2 1 ) 2 − 2 1 ⎭ ⎬ ⎫ + y 2 c l e a r l y , S m i n = − 1
It is short and obvious. Congratulations. Much better then mine. Mine would give only another method of approach.
We can write the given expression as
f(x,y,z) = ( x + 2 y − z ) 2 + ( x + z − 1 ) 2 + y 2 -1
Clearly, for y=0, x=1/2, z=1/2, the square terms become zero.
Thus min. value of f(x,y,z) is − 1
the expression can be as under: K-2 , where K=(x+y)^2+(y-z)^2+(x+y-1)^2+(y-z+1)^2+y^2. Observe that the minimum value of K occurs for y=0 and K= 2x^2-2x+1 +2z^2-2z+1 , so o minimum value of the expression is 1/2+1/2 -2 = -1.
Trying to understand with Latex.
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It must be shown. Differentiate and equate to zero, or give
x-coordinate of the vertex of the parabola OR any other method.
Let f ( x , y , z ) = 2 x 2 + 5 y 2 + 2 z 2 + 4 x y − 4 y z − 2 z − 2 x
At the point of extrema, ∂ x ∂ f = ∂ y ∂ f = ∂ z ∂ f = 0
∂ x ∂ f = 0 ⇒ x + y = 2 1
∂ z ∂ f = 0 ⇒ z − y = 2 1
∂ y ∂ f = 0 ⇒ 5 y + 2 x = 2 z
Solving these 3 equations, we get x = z = 2 1 ; y = 0
Substituting back, we get f ( x , y , z ) = − 1 .
Substituting x = y = z = 0 would give f ( x , y , z ) = 0 proving this extrema is minima.