Someone wanna help me with some math?

[BustAcap]

stupid problem that I can't figure out how to do.. yes, I am dumb!

Given a piece of metal 48" by 60" a square piece will be cut from each corner and the sides folded to form a box. What size piece should be cut to form a box with maximum volume? What will the dimensions of that box be? What size piece should be cut to obtain a volume of 8000 cubic inches?

 

[EvanFerguson]

no
clue

 

[Ameesh]

you in a calculas class?

 

[paruhd0x]

math sux

 

[Whitecloak]

i dont know what grade you are in BustAcap but i assume you know calculus in which case the problem is simple.my assumption is that the box has no top.
assume 'x' is the length of the square piece.
so the length of the box to be constructed will be 60-2x and the breadth will be 48-2x. the height is x
so the volume of the box is x(60-2x)(48-2x)
now differentiate this and set the result to 0 to obtain extrema.
to find the length of the piece to be cut out to obtain 8000 cubic inches, solve the above equation.

 

[thEnEuRoMancER]

Here's the procedure:

Draw a sketch of the sheet metal (rectangle). Draw four squares in corners (that will be cut out). Mark the sides of the square with "a" and the remaining parts of the rectangle (forming the sides of the box) with "b" and "c" respectively.

Then:

2a + b = 60"
2a + c = 48"

b = 60" - 2a
c = 48" - 2a

Volume is:

V = abc = a(60 - 2a)(48 - 2a) = 4a^3 - 216a^2 + 2880a (I didn't double check this!)

Now, in order to obtain the max volume, you need to diferentiate "V" with respect to a and find the zero of the derivative (maximum of the function V(a) ):

dV/da = (edit12a^2 - 432a + 2880 = 0

Solve for "a" for solution to your problem.

 

[Ameesh]

the answer for the first part is 2 * [(21^(1/2)) + 9] which approxzimately = 27.16515

 

[Whitecloak]

Update:
I solved the equation and found that the box will be of max volume when the length of the piece is 8.835inches.
the volume of the box will be 11343 cubic inches.
check my calculations though

 

[BA]

you know how to find the maximum of an equation, right?

________60"_______
|................|
|................|
|................| 48"
|................|
|................|
__________________


cutting out a square'll give you something like this

_______________
_|______________|_
||..............||
||..............||
||..............||
||______________||
|______________|

If the lengths of the sides of those squares are X, you're subtracting 2X from the length and width for the bottom of the box, and if you fold it up, it'll be X high.

 

[pen^2]

a little irrelevant, but i might need some help too.
how do you parametrize 9x^2+6y^2=36 to 3cos(t)i+2sin(t)i=1 <thats the answer, but dunno for sure how to derive it >

 

[Whitecloak]

Ameesh, the length of the square cannot be 27inches coz that is bigger than the original piece. you have to consider the other solution of the quadratic equation which is (18-2*sqrt(21))

 

[pen^2]

cmon guys, help me out a bit

 

[Omegachi]

you guys are smart

 

[Shalmanese]

bah! someone mentions an equation and you all shout &quot;Derive, Derive!&quot;

the x value of the maximum point of a quadratic with eqn ax^2+bx+c = 0 is -b/2a. sub it in to get the y value, x will be the length to cut away or keep depending on how you do the eqn and y will be the area

 

Whoa, you guys rock. Who wants to help me with some linear algebra?