Question

The sum of a number and twice another number is 12. What is their maximum product?

Answer 1

is calculus allowed?

Answer 2

yeah, but i have no idea how to do it

Answer 3

Well, either way we will start with two equations:\[x+2y=12\]and\[xy=P\]I suggest solving the first equation for x and substituting in that expression into the next equation. From there you can either identify the vertex using Algebra formulas, or use differentiation to fin the max value.
Still confused?

Answer 4

find*

Answer 5

no ok i get that, and i have mult choice answers, there 16,18,20,24,and 36, so i can just plug eachof those in?

Answer 6

into what?
No, you should find the answer\[x+2y=12\to x=12-2y\]\[xy=P\to y(12-2y)=12y-2y^2=P(y)\]can you maximize this function?

Answer 7

huh? ok i get the y(12-2y), but then what?

Answer 8

Now use calculus to find the local maximum/minimum of that function.

Answer 9

umm how do i do that?

Answer 10

You've never used calculus before?

Answer 11

nope, i dont think so

Answer 12

In which case you can't use calculus for this problem.

But never mind. Look at P(y) = 12y - 2y^2. That's a quadratic equation. Sketch it using every other tool you have and figure out where the maximum is.

Answer 13

Completing the square would do the job.

Answer 14

Find the vertex of the quadratic equation by completing the square

Answer 15

well how would you do the calculus?

Answer 16

\[P(y)=-2(y^2-6y)=-2(y-3)^2+18\]
What value of y would give you the maximum value for P?

Answer 17

find the derivative of the equation and then set it equal to 0

Answer 18

If you don't calculus, it's not going to help you here. And it isn't something we can explain to you in even an hour in a way you understand.

Follow the procedure Mr.Math just laid out for you.

Answer 19

In the expression I wrote above \((y-3)^2\ ge 0 \implies -2(y-3)^2\le 0\). So the maximum value you would get is at y=3. Right?

Answer 20

\[(y-3)^2\ge 0 \implies \dots\]

Answer 21

oh! is the answer 18?

Answer 22

thanks everyone for helping!!

Answer 23

Yeah, the maximum occurs at y=3 and it's P(3)=0+18=18.

Answer 24

Hey, will all due regards with calculus approach (this statement is specially for James ;) ) I would like to use something more elementary.

A.M \( \ge \) G.M, so, \[ 12 = x+2y \ge 2\times \sqrt{ (x \times 2y)} \Rightarrow 6^2 \ge 2xy \Rightarrow 18 \ge xy\]

which means that the maximum value of the product is \( 18 \).

Answer 25

one number is x, other is
\[12-2x\] and you want to maximize
\[x(12-2x)=12x-2x^2\] vertex of this parabola is at
\[=\frac{b}{2a}=\frac{12}{4}=3\] so you get one numbers is 3, the other number is 6 and the maximum product is \[3\times 6=18\]

Answer 26

@FFM
I doubt that is going to be a familiar theorem to the student.

Answer 27

typo there i meant vertex is at
\[-\frac{b}{2a}=3\]

Answer 28

you certainly need no calculus to find the vertex of a quadratic.

Answer 29

@TuringTest, I don't know, why you think like that since, "Arithmetic mean \( \ge \) geometric mean" is something that is taught in grade school I believe (?!)

Answer 30

I wish, but not to me...