Ed's conic
What is the minimum value of the conic
5 x 2 − 4 x y + y 2 − 1 0 x + 6 y + 5 0
as x and y ranges over all real numbers?
This problem is posed by Ed .
The answer is 40.
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.
5 solutions
This is a beautiful solution with one minor issue: one also has to show that the two squares can be zero at the same time.
The most common incorrect solution was to find the point where both partial derivatives are equal to zero and then claim, without justification, that this is the point of minimum. This is really not obvious. Because you are not allowed to use assume that the problem itself is correct, you cannot just argue that the minimum exists, thus it must be that point.
First, we can factor 5 x 2 − 4 x y + y 2 − 1 0 x + 6 y + 5 0 into 5 ( x − 1 ) 2 + y 2 + ( 6 − 4 x ) y + 4 5 .
By Vieta's formula, we know that 2 a − b is the vertex of any parabola in the quadratic form a x 2 + b x + c .
Since a is positive for y 2 + ( 6 − 4 x ) y + 4 5 , the vertex is a absolute minimum. 2 a − b = 2 4 x − 6 = 2 x − 3 = y .
Now plug y = 2 x − 3 back into the equation. We get 5 ( x − 1 ) 2 + ( 2 x − 3 ) 2 + ( 6 − 4 x ) ( 2 x − 3 ) + 4 5 = x 2 + 2 x + 4 1 .
Taking 2 a − b again, we get x = − 1 as the vertex. ( − 1 ) 2 + 2 ( − 1 ) + 4 1 = 4 0
Therefore, 4 0 is our answer.
If we take the partial derivatives in respect to X and in respect to Y, and equals them to zero, we'll get a system of the maximum or minimum point; in this case is the minimum because both coefficients of X^2 and y^2 are positive.
f(x)=10x-4-10 f(y)=2y-4x+6
10x-4-10=0 2y-4x+6=0
x=-1 y=-5
Now we have the critical points of the function. Substitute the values in the equation:
5(-1)^2-4(-1)(-5)+(-5)^2-10(-1)+6(-5)+50=40
Then we have our value.
Given equation of the conic, 5x^{2} - 4xy + y^{2} - 10x + 6y + 50
Let, p = 5x^{2} - 4xy + y^{2} - 10x + 6y + 50 ---- (1)
Since we have to find the minimum value of the conic, so we are to differentiate the conic w.r.t x and y respectively to obtain two different equations.
Firstly, differentiating both sides w.r.t x ,
\frac{dp}{dx} = 10x - 4y - 10 = 0 ----- (2)
Secondly, differentiating both sides w.r.t y ,
\frac{dp}{dy} = -4x + 2y + 6 = 0 ----- (3)
Note that, both the equations were considered to be equal to zero, so that we can find the values of x & y.
Solving equation (2) and (3),
x = -1
y = -5
Substituting the value of x and y in equation (1), we will get the minimum value of the conic, i.e.
p = 5. (-1)^{2} - 4 .(-1) . (-5) + (-5)^{2} - 10 . (-1) + 6 . (-5) + 50
= 40 [Q.E.D]
The expression given is equivalent to: y 2 + ( − 2 x ) 2 + 2 ( − 2 x ) y + 2 y ⋅ 3 + 2 ( − 2 x ) 3 + 9 + x 2 + 2 x + 1 + 4 0 = ( y − 2 x + 3 ) 2 + ( x + 1 ) 2 + 4 0
The two squares are ≥ 0 , so the minimum value of the expression given is 40