Question

minimize Q=x^2+2y^2, where x+y=3

Answer 1

The first step is to rewrite x+y=3, as y=3-x

Answer 2

kk i got that part

Answer 3

Now we can replace the y in the equation with our newfound value for y, namely y=3-x.

:
Q=x^2+2(3-y)^2
=x^2+18-12x+2x^2
=3x^2+18-12x

Answer 4

Then we have to determine where the minimum of this parabola is

Answer 5

We can do this simply by finding the vertex

Answer 6

do you know how to find the vertex?

Answer 7

yes get the derivative and set it = to 0

Answer 8

x=2 y=1 ?

Answer 9

no

Answer 10

you took a misstep somewhere

Answer 11

Q'(x)= 6x-12

Answer 12

the y value should be 6

Answer 13

no x+y=3 right?

Answer 14

yes

Answer 15

6x-12=0
6x=12
x=2

Answer 16

plug it into the original equation

Answer 17

I did and q=6

Answer 18

not y :)

Answer 19

if x=2 then 2+1=3 y=1
x+y=3

Answer 20

another method is a lagrange multiplier :)

Answer 21

we dont even need anything else, all we need to do is find the vertex

Answer 22

no method was specified

Answer 23

Q=x^2+2y^2; constrained by: x+y=3

F(x,y,L) = x^2+2y^2 -L(x+y-3)

Fx = 2x - L
2x - L = 0
x = L/2

Fy = 4y - L
4y - L = 0
y = L/4

FL = x+y-3
x+y-3 = 0
L/2+L/3-3 = 0
L(1/2+1/3) = 3
L(5/6) = 3
L = 18/5

x = L/2
= 9/5

y = L/4
=9/20

if i did it right

Answer 24

the correct answer was Q=6 x=2 y=1

Answer 25

from the back of the book

Answer 26

hmmm .... guess my lagranging needs more work; either that or i found the maximum :)

Answer 27

lol it was a u shaped parabola

Answer 28

it is: Q=x^2+2(3-x)^2
q=3x^2-12x+18
q'(x)=6x-12
x=2
2+y=3
y=1

Answer 29

(2)^2+ 2(1)^2=6

Answer 30

q=6