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Subject: "Creating a formula"     Previous Topic | Next Topic
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Anthroxy
Member since Jan-15-11
Jan-15-11, 11:40 AM (EST)
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"Creating a formula"
 
   Hi there i saw your post about "Maximum Volume of a Cut Off Box" (https://www.cut-the-knot.org/Curriculum/Calculus/BoxVolume.shtml)and thought that you might be able to help me. I found a similar question on the net and wanted to give it a shot but now i'm completely stuck.It's question b) that i'm having problems with. Well here i go...

Question:

a) We have a folded box shaped like a rectangle with the length b and width a. The box has four squares that we fold up, these squares are the height of the box x. Choose a couple of values for length b and width a and decide what x has to be in order for the box to have the largest volume.

b) Produce a formula that enables you to calculate what x has to be in order to get the largest volume when only given the length a and width b.

My attempt to a solution:

a)

1. First i got to make a function that describes the volume of the box
2. Derive the function in order to find the maximum point of the function
3. Find out which x-values that can't be a part of the solution (this is where my first problem lies) for example that x > 0 but can't be larger than...
4. If both x values are a part of the solution, then find out which one lies at a maximum point and finding the y coordinate by inserting the x value into the normal function.

Length: b-2x
Width: a-2x
Height: x

Attempt 1:

y= x(b-2x*a-2x)
y= x(ab + b*-2x + a*-2x +4x^2)

I then chose a = 20 b = 5 (measured in cm) this gave me

y= x(20*5 +5*-2x -20*-2x +4x^2)
y= 4x^3 -50x^2 +100x
y´= 12x^2 -100x +100
y´= x^2 - 25/3x + 25/3

When solving the equation using the quadratic formula i get x= 1,16 and x= 7,16 and when searching for the maximum point i get that when x= 1,16 we're at a maximum point which gives us y = 54,94. The maximum volume being 54,94cm^3 when x = 1,16.

Attempt 2:

a= 10
b=5

On attempt 2 i got x = 3,94 and x= 1,05 and the maximum point was at x= 1,05 and y= 54,5

Attempt 3:

a= 40
b= 20

On attempt 3 i got x= 15,77 x= 4,22 and the maximum point was at x= 4,22 and y= 1539,59.

b) This is where my problem lies, creating a formula where i only need a and b to get what x needs to be in order to get the maximum volume

Thanks in advance Mathmen


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alexbadmin
Charter Member
2712 posts
Jan-15-11, 11:56 AM (EST)
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2. "RE: Creating a formula"
In response to message #0
 
   You have actually done everything yourself, except that your algebraic notations lacks consistency and do not follow the commonly accepted rules, e.g.
y= x(b-2x*a-2x)

is awfully ambiguous. If you did not know where you came from you would not be able to interpret the expression correctly. The right way should be

y= x(b-2x)*(a-2x)

What you wrote has a completely different meaning. The next one

y= x(ab + b*-2x + a*-2x +4x^2)

I would write either as

y= x(ab + b*(-2x) + a*(-2x) +4x^2)

or as

y= x(ab - 2bx - 2ax +4x^2).

So this is your function (get it by multiplying out):

y = 4x³ - 2(a + b)x² + abx.

To find the extremes you differentiate and equate to 0:

y' = 12x² - 4(a + b)x + ab = 0.

This is a quadratic equation with roots

x = {4(a + b) ± sqrt(16(a + b)² - 48ab)} / 24

Simplify:

x = {(a + b) ± sqrt(a˛ - ab + b˛)} / 6

For you to determine which of these is a maximum.



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