Differential calculus to determine maxima and minima

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corn_she

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Use differential calculus to determine the minima values of a required volume.

A rectangular sheet of metal having dimensions 30cm by 18 cm has the squares removed from each side of the four corners and the sides bent upwards to form an open box.

Determine the maximum volume of the box, including the correct SI units.


Please could you give/show full working.
 
corn_she said:
Use differential calculus to determine the minima values of a required volume.

A rectangular sheet of metal having dimensions 30cm by 18 cm has the squares removed from each side of the four corners and the sides bent upwards to form an open box.

Determine the maximum volume of the box, including the correct SI units.


Please could you give/show full working.
Click to expand...

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting
 
I have only just starting learning this.....so i'm stuck straight from the beginning.
 
corn_she said:
I have only just starting learning this.....so i'm stuck straight from the beginning.
Click to expand...

Then first make a sketch of the rectangular sheet

Now sketch four squares at four corners - each square has sides of 'x' length.

If you cut-out these squares - you'll be able to make an open-top box - by folding the sheet.

What will be the volume of this box?
 
I need a height of the box to find the volume

Don't i need to find out how big the squares are at each corner?
 
corn_she said:
I need a height of the box to find the volume

Don't i need to find out how big the squares are at each corner?
Click to expand...

That is what you will calculate from maximizing the volume. Right now assume the squares to be x by x.

The height of the box will be 'x'.
 
corn_she said:
Please could you give/show full working.
Click to expand...
Sorry, but this site doesn't do "cheetz". :shock:

corn_she said:
A rectangular sheet of metal having dimensions 30cm by 18 cm has the squares removed from each side of the four corners and the sides bent upwards to form an open box.
Click to expand...
Start with what you learned back in algebra:

Draw the rectangular sheet. Label the given dimensions. Draw the four squares at the corners. Label their (unknown) dimensions with a variable. (Since you'll be folding up the "sides" created by these square corners, the dimensions of these squares will create the height. So let's use "h" for the variable.) Draw dashed lines showing where the sheet will be folded up.

Subtracting what you just folded up, what are the dimensions of the base of the box, in terms of the variable? Use these "length" and "width" dimensions, together with the "height" dimension, to create an expression for the volume V of the box, using the basic volume formula you memorized back in algebra.

You can now create a function for the volume V in terms of the variable. Differentiate, and maximize, like they've taught you in calculus.

If you get stuck, please reply showing all of your steps in following the step-by-step instructions above. Thank you! ;)
 
I havent got a clue.......you need to remember i'm a 35 year guy trying to learn maths again. it's been a long time since ive been a school....
 
corn_she said:
I havent got a clue.......you need to remember i'm a 35 year guy trying to learn maths again. it's been a long time since ive been a school....
Click to expand...
Come on, you have to try. Take a sheet of paper and imagine it is 30cm by 18cm. Now go to each corner and cut out a small square of the same size. I would say that the dimension of the square is xcm by xcm but a previous poster requested that you call it hcm by hcm.
Now 100% for real you need to fold this paper on your own so that you get a box. We know the height of the box is h. The length originally was 30cm but we made it smaller, so what is the length now?. The width originally was 18cm but is now smaller. What is the width now? The volume is V=L*H*W. L, H and W are in terms of h.
You were asked to do this a few times now. I promise you if you make this box on your own you will figure out the volume. Let us know what you get.
 
Ive cut the box out...

v=l.w.h
length = 30-2x
width = 18-2x
h= x

v(x) = X (30-2x)(18-2x)

Is that right?
 
corn_she said:
Ive cut the box out...

v=l.w.h
length = 30-2x
width = 18-2x
h= x

v(x) = X (30-2x)(18-2x)

Is that right?
Click to expand...
Perfect. I knew you could do it if you tried. To minimize the volume you need to take the derivative of V and set it equal to 0 and solve for x.
 
corn_she said:
28x16x2= 896cm^3
Click to expand...
Show your work. Do you really think that someone is going to work out the problem and see if your result is correct? Is that fair of you?
Actually your answer has to be wrong since if x=2 then L and W would NOT be 28 and 16. So try again.
 
V(x) = x(30-2x)(18-2x)
V(x) = x * (540-60x-36x+4x2) ....................(1)
= x * (4x2 - 96x + 540)
= 4x3 - 96x2 + 540x

V1(x) =12x2 - 192x + 540

For maximum/minimum, V1(x) = 0 →

12x2 - 192x + 540 = 0

x2 - 16x + 45 = 0

x1,2 = 8 ± √(64-45) = 8 ± 4.358 = 12.358 or 3.641


X= 192sqauaredroot 36864 - 4.12.540/24
=296.6135746/24
=12.35889894

X= 192-sqauaredroot 36864 - 4.12.540/24
=87.38642536/24
=3.641101056

= 12.359 or 3.641

Answer is 3.641

What do i do now?
 
Last edited by a moderator:
corn_she said:
V(x) = x(30-2x)(18-2x)....................(1)
V(x) = x * (540-60x-36x+4x2)
= x * (4x2 - 96x + 540)
= 4x3 - 96x2 + 540x

V1(x) =12x2 - 192x + 540

For maximum/minimum, V1(x) = 0 →

12x2 - 192x + 540 = 0

x2 - 16x + 45 = 0

x1,2 = 8 ± √(64-45) = 8 ± 4.358 = 12.358 or 3.641


X= 192sqauaredroot 36864 - 4.12.540/24
=296.6135746/24
=12.35889894

X= 192-sqauaredroot 36864 - 4.12.540/24
=87.38642536/24
=3.641101056

= 12.359 or 3.641

Answer is 3.641

What do i do now?
Click to expand...

You can see that the value of x cannot be 12+ (Why?)

Use x = 3.641 in (1) and calculate V for answer

Do not forget to include the correct dimension (in SI units) for volume.
 
Last edited by a moderator:
it cant be 12 because:

18-2x>0
18>2x
9>x
0> <9

Volume = v(x) = (30-2 *3.641)(18-2*3.641)
=243.491524 cm^3
 
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