Question

I need help minimizing Q=3x^2+y^2, where x+y=4

Answer 1

write Q in terms of one variable using the equation x+y=4

Answer 2

is this a calculus or algebra question?

Answer 3

Calc

Answer 4

I'm confused

Answer 5

so you know to minimize or to maximize this involves finding the derivative of thing you wish to minimize or maximize ?

Answer 6

Please help me step by step

Answer 7

Do you know how to use x+y=4 to write Q in terms of one variable ?

Answer 8

If not, here is a hint: Solve x+y=4 for one of the variables

Answer 9

No i don't

Answer 10

There is no wrong or right variable to solve for in x+y=4
just pick one and solve for it

Answer 11

Plug in any number ?

Answer 12

no solve x+y=4 for x

or

solve x+y=4 for y

Answer 13

don't do both
just do one of those things

Answer 14

you only need one step to solve either equation

Answer 15

try subtracting something on both sides

Answer 16

I'm lost

Answer 17

x+y=4
Say we want to solve for y
notice you have a plus x next to it
to undo an addition of x on that side with the y
you will need to subtract x on both sides

Answer 18

what equation do we have after doing that ?

Answer 19

\[x+y=4 \\ (x+y)=4 \\ \\ (x+y)\color{red}{-x}=4\color{red}{-x} ... \\ \text{ what do we have after subtracting } \\ x \text{ on both sides } \\ \text{ what does the left hand side simplify to }\]

Answer 20

Y=4-X

Answer 21

right
now write Q in terms of x only

Answer 22

Q=4((Y-6)Y+12) ??

Answer 23

IS that the answer?

Answer 24

not entirely sure I see how you got that

Answer 25

didn't you say \[Q=3x^2+y^2 \text{ and } y=4-x\]

Answer 26

Yeah

Answer 27

just replace y with (4-x)
to write Q in terms of one variable

Answer 28

So now I would be solving Q=3x^2+(4-x)^2 ?

Answer 29

You differentiate that
and set Q'=0 and solve for x to find critical numbers

Answer 30

So the x values are 0?

Answer 31

no solving Q'=0 for x gives the values of x for which the graph of Q has horizontal tangents and could possibly tell us where we have max or min if any

Answer 32

How would I plug this into a graphing calculator?

Answer 33

Q is actually a parabola which is easy to find max/min of even without calculus
this is why I asked if it was an algebra or calculus question

but either way is easy
i honestly like derivatives more :p

parabolas have min/max occur at the vertex... and the way to determine if the vertex is a max/min is to determine if the parabola is open up or down and you can do this by looking at the coefficient of x^2


If this is a calculus question though, I suggest using calculus ways and differentiate Q
you need power rule, constant rule, constant multiple rule, and you could also use chain rule if you don't expand that (4-x)^2 part

Answer 34

\[\frac{d}{dx}(3x^2)=? \\ \frac{d}{dx}(4-x)^2= ?\]

Answer 35

Idk how to do this
This is not helping