Question

Solve the given optimization problem by using substitution.

Find the minimum value of
f(x, y, z) = 2x^(2) + 2x + y^(2) − y + z^(2) − z − 5
subject to z = 2y.

I actually tried to do this one and couldn't figure it out

Answer 1

Hmm, don't you substitute \(2y\) for z throughout \(f(x,y,z)\) giving you\[f(x,y,z) =2x^2+2x+y^2-y+(2y)^2-(2y)-5 = 2x^2+2x+5y^2-3y-5\]
Then take the partial derivatives of that with respect to \(x\) and \(y\) and set them equal to 0? Solve those two equations and back substitute to find \(z\). Isn't that how you work these?

Answer 2

yeah but how do you find the minimum

Answer 3

The point at which it is a minimum, or the value there?

Answer 4

the value

Answer 5

Just plug in the values. Do you have the values of \(x,y,z\) yet?

Answer 6

i have x as -1/2 and y as 1/3

Answer 7

y as 1/2

Answer 8

x = -1/2, y = 3/10 are what I got. z = 2y = 6/10 = 3/5

Answer 9

\[f(x,y,z) = 2x^2+2x+y^2-y+z^2-z-5\]Substitute \(z=2y\)\[f(x,y) = 2x^2+2x+y^2-y+(2y)^2-(2y)-5 \]\[f(x,y)= 2x^2+2x+5y^2-3y-5\]

Agreed so far?

Answer 10

yeah that makes sense so far

Answer 11

\[\frac{\partial \left(f(x,y)\right)}{\partial x} = 4x+2\]
\[\frac{\partial \left(f(x,y)\right)}{\partial y} = 10y-3\]

Answer 12

Setting those equal to 0 and solving, we get
\[4x+2=0\]\[4x=-2\]\[x=-\frac{1}{2}\]
\[10y-3=0\]\[10y=3\]\[y=\frac{3}{10}\]\[z=2y\]\[z=2*\frac{3}{10} = \frac{3}{5}\]
Agreed?

Answer 13

okay yeah i got that, I'm still a little confused on how you got 5y^(2)

Answer 14

Oh, remember \(z = 2y\) so \(z^2 = (2y)^2 = 4y^2\)

Answer 15

i keep getting 3y^(2)

Answer 16

and there's another \(y^2\) from the first part of the equation to add to that...
I made the same mistake :-)

Answer 17

wouldn't that make it 4y though? not 4y^(2)?

Answer 18

if \(z = 2y\), \(z^2 = z*z = (2y)(2y) = 2*2*y*y = 4y^2\)

Answer 19

ohhh gotcha haha my bad

Answer 20

so then how do you find the minimum value?

Answer 21

yeah, that's why I try to remember to put () around things when substituting...

Answer 22

\[f(x,y,z) = 2x^2 + 2x+y^2-y+z^2-z-5\]Substitute in our values of \(x,y,z\)
\[f(-\frac{1}{2},\frac{3}{10},\frac{6}{10}) = 2(-\frac{1}{2})^2 + 2(-\frac{1}{2}) + (\frac{3}{10})^2-(\frac{3}{10})+(\frac{6}{10})^2-(\frac{6}{10}) - 5\]\]\[=2(\frac{1}{4})-1+\frac{9}{100}-\frac{30}{100}+\frac{36}{100}-\frac{60}{100}-5\]\[=-6+\frac{1}{2}-\frac{45}{100} = -\frac{119}{20} = -5.95\]

Answer 23

Here's a contour plot:

You can see that \((-\frac{1}{2},\frac{3}{10})\) is right in the middle of the innermost contour line.

Answer 24

oooh geez thats why i couldn't figure this one out but thank you so much!!! i really appreciate you showing the steps

Answer 25

If you have the patience (or a programmable calculator), you can crank out some values nearby and verify that this is a minimum, not a maximum.

Answer 26

Would you believe this is the first time I've solved one of these? :-)

Answer 27

haha no way?!

Answer 28

Here's a somewhat zoomed in contour plot, in color.

Answer 29

well you did an awesome job :)

Answer 30

thats the graph of it?

Answer 31

and do you possibly have time to help me with one more? :) its a different type of problem i just need an example to see how to do it

Answer 32

Gave me an excuse to experiment with Mathematica's contour plotting :-)

Yes, lighter colors are larger values, darker colors are smaller. It's like a topographical map of the surface of the function — everything on one of those rings has the same value of \(f(x,y,z)\)

Answer 33

Well, there's one way to find out! :-) Post the new question and tag me...

Answer 34

how do i tag lol

Answer 35

just put "@whpalmer4" in a post

Answer 36

okay gotcha thank you so much! :)

Answer 37

then I get a notification and a link I can click to go there without having to search for it.