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# algebrA

+10
248
9
+1162

Find the minimum value of x^2+2xy+2y^2+4y+1 over all real numbers x and y.

Jul 4, 2022

#1
-2

Let f(x,y) = x^2 + 2xy + 2y^2 + 4y + 1 = 0

Taking partial derivatives, we get

2x + 2y = 0

2x + 8y + 4 = 0

Solving the system, we get  x = 2/3 and y = -2/3.  So the minimum value is f(2/3,-2/3) = -11/9.

Jul 4, 2022
#2
+1162
+9

sry, but it's incorrect.

nerdiest  Jul 4, 2022
edited by nerdiest  Jul 5, 2022
#3
+3

Hi!

One way: (Using differentiation).

Find the minimum of:          \(x^2+2xy+y^2+4y+1\)

Let \(F(x,y)=x^2+2xy+y^2+4y+1\)

\(F_{x}=2x+2y\)             (Partial derivative with respect to x)

\(F_{y}=2x+4y+4\)      (Partial derivative with respect to y)

We set these equal to zero:

\(x+y=0\)

\(x+2y=-2\)

Solve the system of two equations:    (By subtracting the first equation from the second.)

\(y=-2 \implies x=2\)

Therefore, the minimum occurs when: \(x=2,y=-2\)

The minimum is: \((2)^2+2(2)(-2)+2(-2)^2+4(-2)+1=4-8+8-8+1=-3\)

So the minimum value is -3.

Second way: (Uses completing the square and the fact that \(a^2\ge 0\) for any real a.)

\(x^2+2xy+2y^2+4y+1\)

Consider: \((x^2+2xy+y^2)+(y^2+4y+1)\) , complete the square on each bracket:

\((x+y)^2+y^2+4y+1=(x+y)^2+(y+2)^2-3\)

Notice: \((x+y)^2 \ge 0 , (y+2)^2 \ge 0\) for all x and y.

So,

\((x+y)^2+(y+2)^2\ge 0 \iff (x+y)^2+(y+2)^2-3 \ge -3\)

The L.H.S is what we want to find the minimum of.

Therefore, the minimum of the desired expression is -3.

Hope this helps!

Jul 5, 2022
#4
+1162
+9

but I will give you credit for your hard work.

nerdiest  Jul 5, 2022
#5
+1162
+9

oh no... i answered the question wroooong! NOOOOOO. You in fact might be correct. I'm the one who is dummmmmb. dang it.

nerdiest  Jul 5, 2022
#6
+2

Nah, It's ok to make mistakes, the important thing is to learn! :)

Guest Jul 5, 2022
#7
+1162
+9

yes

uuuum. random question, but are you just a guest, or a member logged out.

nerdiest  Jul 5, 2022
#8
+2

guest.

Guest Jul 5, 2022
#9
+1162
+7