Question

Maximizing area within a constraint

I've got this problem (picture below) and I'm kind of stuck. Here's what I have so far:

let x be length and y be height. x + y must equal (6 - x) / 2 to give the greatest possible area. To find area, use x * y. I got:

A(x) = x * (6 - x) / 2 = 1/2(6x - x^2)

A'(x) = 3 - (x / 2)

So I found that A'(x) = 0 when x = 3. But for one thing, when I try to find height (y) by substituting 3 into x + y = (6 - x) / 2, I get -3/2, which can't be right. And for another, I don't know if my process is right at all.

Any help is appreciated.

Answer 1

Woops sorry, A'(x) = 0 when x = 6. But that would still give me an answer of 0 when substituting 6 for x into x + y = (6 - x) / 2

Answer 2

where is x+y=(6-x)/2 given?

Answer 3

I see (x,y)=(x,(6-x)/2)
so length is x
and height is (6-x)/2

Answer 4

area of rectangle=length*height
so you have area=x*(6-x)/2

Answer 5

so you find area'
then set area'=0 and solve for x

Answer 6

I think you sorta did that

Answer 7

area=(6x-x^2)/2
area=3x-x^2/2
area'=3-x
isn't it?

Answer 8

So area'=0
when 3-x=0
which is when x=?

Answer 9

I'm sorry, I didn't mention: x+y = (6 - x) / 2 came from me. I reasoned that myself.

Yup, area is 3x - (x^2 / 2) so area' would be = x - 3

Answer 10

area'=3-x

Answer 11

but yeah that yields the same critical number

Answer 12

3-x=0
when x=3
then you can find the height=y=(6-x)/2

Answer 13

replace x with 3

Answer 14

then to find the area just multiply your height and your length

Answer 15

weird thing is your picture sorta hints at the answer

Answer 16

Oh derp. Hold on... I'm slow, gotta process yo

Answer 17

if i were you I would totally disregard the equation x+y=(6-x)/2

and look at (x,y) is given as (x,[6-x]/2)
Area=x*y
=x*(6-x)/2
=(6x-x^2)/2
=3x-x^2/2
Area'=3-x
Area'=0 when x=3
So x=3
and y=(6-x)/2 when x=3
Then the max area is given by x*y
so you are almost finished here :)

Answer 18

You're totally right. I was getting confused because every time I would do this question, I switched between area' = (6 - 2x) / 2 and area ' = 6 - x/2 because I kept forgetting the derivative of x^2 is 2x and not x.

Anyway, area' = 0 when x = 3. So y = (6 - 3) / 2 = 3/2 and the area of the rectangle would be 3 * 3/2 = 9/2

Answer 19

derivative of x^2 is 2x
but the derivative of x^2/2 is 2x/2=x

Answer 20

and yes your area looks fabulous

Answer 21

and so do your dimensions that give the max area

Answer 22

Thank you freckles :) It's always the small details that get me.

Answer 23

\[A=\frac{6x-x^2}{2} \\ A=3x-\frac{x^2}{2} \\ A'=3-\frac{1}{2}(2x) \\ A'=3-x\]
just for fun I want to be sure you got this for A'
which I'm certain you did because you have the right dimensions and right area

Answer 24

Great cross checking \(\checkmark\)

Answer 25

Yup, that's what I got. I scratched this question on paper a few times prior to asking here and I did this half the time:

\[A = \frac{6x - x^2}{2}\]

\[A' = \frac{6 - x}{2}\]

\[A' = 3 - \frac{x}{2}\]

And the other half of the time I did it correctly.

Answer 26

something looks wrong here.

Answer 27

i think that is the part he is scratching off his paper

Answer 28

Oh yeah, gotcha :D

Answer 29

Yeah, freckles is correct :)

Answer 30

I think I've got the answer in order now. Thank you freckles! And also thank you, Jhannybean :)