Question 201030: not sure if in right area but can someone help me with this problem
An open box is to be constructed from a piece of cardboard that is 30in. by 30in. by cutting a square out of each corner and folding up the sides. What are the dimensions of the box that will yield the maximum volume?
so confused
thanks
Found 3 solutions by solver91311, stanbon, RAY100:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
There is no correct area on this site for this problem because it is a Calculus local maximum problem. The closest thing I could find was Algebra: Polynomials, rational expressions, and equations
Let  represent the measure of the side of the squares cut out of the corners. That means that the width and length of the resulting box must measure  and the height of the resulting box is just . So you can write a function representing the volume of the box in terms of the dimension of the cutout square thus:
 = x(30 - 2x)(30 - 2x) = 4x^3 - 120x^2 + 900x)
Given a continuous and differentiable function, such as the above polynomial function, local extrema can be found at the value of the independent variable for which the value of the first derivative is equal to zero. So, take the first derivative:
Set the first derivative equal to zero and solve:
(4x - 20) = 0)
Hence, the -coordinates of the local extrema are 5 and 15.
Given a continuous and twice-differentiable function if the 2nd derivative is less than zero evaluated at the -coordinate of an extrema, that extrema is a maximum. If it is positive, the extrema is a minimum.
Take the 2nd derivative:
Then evaluate at 5:
 = 24(5) - 240 = -120)
So 5 is the -coordinate of a maximum.
 = 24(15) - 240 = 120)
So 15 is the -coordinate of a minimum. Which, by the way, makes very good sense because if you cut 15 inch squares out of the corners of a 30 by 30 piece of material, you would end up with a 15 X 0 X 0 box having zero volume.
So, you cut 5 inch squares from each corner, leaving you with a 5 X 20 X 20 box.
Here's a picture of your volume function:
John
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! An open box is to be constructed from a piece of cardboard that is 30in. by 30in. by cutting a square out of each corner and folding up the sides. What are the dimensions of the box that will yield the maximum volume?
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Draw the 30 by 30 piece.
Draw a square in each of the four corners.
Let the dimensions of each square be "x" by "x".
Imagine cutting out those four squares and folding
up the remaining flaps to make a box with no top.
--------------
The base is "30-2x" by "30-2x" and the height is "x".
------------------
The volume is as follows:
V = x(30-2x)^2
V(x) = 900x - 120x^2 + 4x^3
-----------
Take the derivative and set it equal to zero:
V'(x) = 900 - 240x + 12x^2 = 0
Sove for "x":
2x^2 - 40x + 150 = 0
(x-15)(2x-10)= 0
x = 15 or x = 5
---
x = 15 gives you a minumum volume of zero.
x = 5 gives you a maximum volume of 5*20 = 100
====================================================
Cheers,
Stan H.
Answer by RAY100(1637) (Show Source):
You can put this solution on YOUR website! A rough sketch of the net will help.
It is a square, 30 x 30, with square cutouts at each corner.
The cutouts are x by x.
This makes the base (30-2x) square
.
The volume of the folded box is ,,,Vol = L*W*h = (30-2x)(30-2x)(x)
.
V = {900 -120x +4x^2} *x
.
V= 4x^3 -120x^2 +900x
.
a table of options
.
x,,,,,,,,,,Volume
0,,,,,,,,,0
1,,,,,,,784
2,,,,,,1352
3,,,,,,1728
4,,,,,,1936
4.9,,,,1999
5,,,,,,2000,,,,,,max
5.1,,,,1999
6,,,,,,1994
10,,,,,1000
15,,,,,,,,0
.
Therefore max box is 20 x 20 x 5 =2000 cubic inches
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