Question

how to prove that a rectangular box with square base and top having a fixed surface area and maximum volume is a cube

Answer 1

This is an optimization problem

Answer 2

Do you know how to tackle those?

Answer 3

if is a cube the V= x^3, A= 2X^2 + 4X^2, derivative A'= 12X so 12X=0, X=0

Answer 4

Ok, the question says that the S.A. is fixed, correct? So let's give it a S.A. of 500

a "square-based rectangular box" has length the same as width (to keep the base square). This gives you these equations, with w (width and length) and h (height) of the box:

surface area (square top/bottom, plus four sides):
a = 2w^2 + 4wh
volume:
v = w^2h

Do you follow that?

Answer 5

yes, i was x's and u r using w,h

Answer 6

got it

Answer 7

Here is the picture to show you visually what I was saying.

Now we have our two equations, lets plug in what we know

\[\Large 500=2w^2+4wh\]
&
\[\Large A=w^2h\]

Answer 8

We are trying to MAXimize Area, so we have to find the max of that function (we'll use derivative in a moment to find it). But Area is in terms of w and h, we need to get it into terms of just w or just h, which one do you want to get it in?

Answer 9

V= W^2h right?

Answer 10

Correct.

Answer 11

We have to get it in terms of one variable, we have the Surface area equation to help us so we have to

1) Isolate either h or w in the surface area equation.
2) Plug it into the area equation

So do you want to isolate w or h?

Answer 12

h= v/w^2

Answer 13

I advise we go along with h, because I solved it isolating both w and h and h was easier

Answer 14

We have to isolate w or h in the SURFACE AREA equation and plug it into the **volume (I said area before sorry) formula

Answer 15

got it, what is the next step?

Answer 16

We isolate it, can you tell me what you got?

Answer 17

i isolated h and plugged in V

Answer 18

Ok, can you tell me what your new Volume formula is?

Answer 19

V=w^2*(500-2W^2/4w)

Answer 20

Alright, I'll clean it up for you to

\[\Large V= {500w-2w^3 \over 4}\]

Answer 21

Now we have to take the derivative of that

Answer 22

thanks

Answer 23

Tell me what you get

Answer 24

v'=(200-24w^2)/16

Answer 25

2000

Answer 26

remember that we can factor out the 1/4 when we take the drivative and then we multiply our answer by 1/4 after \[\Large {dy \over dx}[c*f(x)]=c*f \prime x\]

Answer 27

So it's really the derivative of \[\Large 500w-2w^3\]
and then take the answer and put it over 4

Answer 28

\[\Large {500-6w^2 \over 4}\]

Answer 29

what about this 250 - 3w^2 / 2

Answer 30

Tell me how you came up with that result and maybe I can help you with what you're doing wrong

Answer 31

i just simplified by 2

Answer 32

Oh, lol I didn't see that, yes those two are the same thing

Answer 33

\[\Large {250-3w^2 \over2}=0\]

Answer 34

Solve for x (leave it in radical form)

Answer 35

w*******

Answer 36

w=250/3^1/3

Answer 37

\[\sqrt[3]{250/3}\]

Answer 38

Ok, it's \[\Large \sqrt{{250 \over 3}}\] sqrt not cuberoot. Now we're going to plug in a point less than that into our derivative equation and then plug in a number greater than that inot our derivative equation to see a max

Answer 39

point less such as \[\sqrt{250/4}\]