What are the x & y values at the minimum?

Question

Find the x and y values at the minimum of
f(x)=2x²+xy+2y²+6y+7

 

I have tried this multiple times but I keep getting marked wrong. I'm not really sure how to approach this correctly so it would be super helpful if work could be written out for me! Thank you! 

 

(university level math)

Answer 1

f(x)=2*x²+x*y+2*y²+6*y+7

 

      = (2*x² + x*y) + 2*y² + 6*y + 7

 

      = 2*(x² + x*y/2) + 2*y² + 6*y + 7

 

      = 2*(x + y/4)² - y²/8 + 2*y² + 6*y + 7 = 2*(x + y/4)² + (15/8)*y² + 6*y + 7

 

      = 2*(x + y/4)² + (15/8)* (y² + (16/5)*y) + 7

 

      = 2*(x + y/4)² + (15/8)* (y + 8/5)² - (24/5) + 7

 

     = 2*(x + y/4)² + (15/8)* (y + 8/5)² + (11/5)

 

∴ Since f(x) is the same as the sum of two squares and a constant term, if the values of the two squares can be made zero then f(x) has a minimum that is the same as the constant term.

 

The square terms become zero when (x + y/4) = 0 and (y + 8/5) = 0

 

∴ y = -8/5 = -1.6 and x = -y/4 = 2/5 = 0.4

   The minimum value of f(x) = 11/5 = 2.2

 

 

Answer 2

So far to my knowledge

 

If we take first order condition i.e. (FOC)

 

we get

 

df(x)/dx = 4x+y=0

 

which means, y= -4x           equation 1

 

again

 

df(x)/dy = x+4y+6=0          equation 2

 

Putting the value of y in equation 2 we get

 

x+ 4*(-4x)+6=0

x-16x=-6

-15x=-6

x= -6/-15

x= 2/5

 

Now, putting the value of x in equation 1, we get

 

y= 4x

y = 4*2/5

y= 8/5

 

Therefore, the value of x and y are 2/5 and 8/5 respectively