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[SOLVED] max dimensions of box template

karush

Well-known member
Jan 31, 2012
2,719
find the dimensions of the largest rectangular box with a square base and no top that can be made from out of $1225 in^2$ of material.



well, this might be over simplified but since the volume is max of cube to the a surface area I thot making a cube out of the material would be max volume assuming that is what they mean by largest rectangular box.

since the $\displaystyle\sqrt{1225} = 35$ then $\frac{35}{3}$ would be the length of the side of the cube and to cut the material accordingly. $x$ being the side.

View attachment 525

I thot to that usually one can get a max value with a parabola but the vertex of $9x^2$ is $0$ there is no book answer for this...mostly making sure I understand the question correctly..... thx ahead...
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
EDIT: I don't think that this interpretation of the problem is correct. It should be made from one sheet of material that is cut and folded up to make the box. The diagram you drew has pieces cut out from it, so the below assumes your interpretation but I'm not sure.

Max/min problems like this always follow the steps of first writing an expression with 2 variables and then making a substitution to get it to just 1 variable. From that point you then differentiate and find the solution.

Let's say the square base has length x and width x. The height of the box is unknown so let's call that y. You are also correct that we need to use surface area now. The total surface area must be 1225, so let's express that in terms of x and y. There is the base and 4 sides of the box. The base uses $x^2$ square inches and each side uses $xy$ so all 4 use $4xy$. That brings the total to $x^2+4xy=1225$

Solving that for y we get \(\displaystyle y=\frac{x^2-1225}{4x}\). Now we look at the volume of the box $x^2y$ and plug in the previous equation for y to get the whole thing in terms of 1 variable.

See where that takes you.

EDIT2: This is supposed to use calculus I believe. You were given this problem in your calc course, correct?
 

karush

Well-known member
Jan 31, 2012
2,719
so you are saying that

$\displaystyle x^2\left(\frac{x^2-1225}{4x}\right)= \frac{x^3-1225}{4}=V$

when that graphed as a cubic function the rel max is little over $4000$

actually yes I am in calc but this was given as an algebra 2 problem just someone I am trying to help, but didn't know how to put this in algebra terms. So apparently this may not be a cube for max V

I know that the y' = 0 will give min/max
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
EDIT2: This is supposed to use calculus I believe.
True, unless you, karush, happen to be very good at maximizing cubic polynomials without calculus. Incidentally, to finish this problem, I think you're going to need the boundary limits, because taking the derivative $dV/dx$ and setting it equal to zero finds a nice minimum.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
so you are saying that

$\displaystyle x^2\left(\frac{x^2-1225}{4x}\right)= \frac{x^3-1225}{4}=V$
Careful. I think the $1225$ needs an $x$ multiplier.
 

karush

Well-known member
Jan 31, 2012
2,719
$\displaystyle x^2\left(\frac{x^2-1225}{4x}\right)$=
$\displaystyle =\frac{x^3-1225x}{4}=V$

yep saw that, thanks,

well the only way I know to find max with algebra
is to find midpoints between the zeros
then find $f(x)$ if it is + then it is max
kinda messy process
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
Well, as I said in Post # 4, taking the derivative and setting it equal to zero finds, in this case, a local minimum. To find the max, you're going to need to find some sort of boundary conditions, and try to evaluate the volume at the extremes.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
Well, as I said in Post # 4, taking the derivative and setting it equal to zero finds, in this case, a local minimum. To find the max, you're going to need to find some sort of boundary conditions, and try to evaluate the volume at the extremes.
Yep, my mistake earlier. The maximum volume is desired.

For the boundary condition, use the fact that \(\displaystyle x^2 \le x^2+4xy\) assuming $x \ge 0$ and using $x^2+4xy=1225$.
 

karush

Well-known member
Jan 31, 2012
2,719
whoa lost you there not familiar with boundary conditions....

y' I got $\frac{3x^2}{4}-\frac{1225}{4}$
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
whoa lost you there not familiar with boundary conditions....

y' I got $\frac{3x^2}{4}-\frac{1225}{4}$
What happens when you set that equal to 0? What do you find? A local maximum or local minimum?
 

karush

Well-known member
Jan 31, 2012
2,719
we are left with $-\frac{1225}{4}$ which is as mentioned a min but the cubic graph also shows a max at around 4125
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
we are left with $-\frac{1225}{4}$ which is as mentioned a min but the cubic graph also shows a max at around 4125
Correct, and that's a problem. If x = 4125 then when you plug that into the surface area formula you see that you don't have enough material. So the min/max method won't work here. You need to set up boundaries.

Because $x>0$, \(\displaystyle x^2 \le x^2+4xy\) and the part on the right hand side is 1225 so that's the same as \(\displaystyle x^2 \le 1225 \rightarrow x \le 35\).

So in general, $0 \le x \le 35$. What do we see about the graph of the volume function in this domain?
 

karush

Well-known member
Jan 31, 2012
2,719
we are left with

$-\frac{1225}{4}$ which is min

however are we including the cutout of corners of a fixed square of material
which will alter the SA
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
we are left with

$-\frac{1225}{4}$ which is min

however are we including the cutout of corners of a fixed square of material
which will alter the SA
I assumed your interpretation of the problem was that this wasn't cut out from anything. That's why the surface area is $x^2+4xy$. In real world applications you don't have access to shapes like this without cutting something out so often this problem is given with the assumption you have to cut a rectangular sheet to make the box.

There is something I seem to have missed because on the domain of [0,35] the volume is negative so none of those values work. Maybe someone else can jump in a figure out what's going on?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
I suggest that karush post the original problem exactly as given, word-for-word. We can go from there.
 

karush

Well-known member
Jan 31, 2012
2,719
OK, could we close this thread, I will restate the original question in a new post, although looks like what I have is correct in the original question of the OP. basically we are making a box out of 1225 sq in of material and looking max volume. at least that what it seems to be asking.,

sorry, didn't mean to take everyone for a walk in the woods...(Movie)