Question

An open box is to be formed from a square sheet of carboard (square is 10x10 cm) by cutting squares and then folding up the sides. (the squares cut off are just the corners as they are labeles as an x by x).
A) Find a function for the volume of the box. State the domain of your function.
B)Find the value of x that maximizes the volume.

Answer 1

Base of the box would be (10-2x) by (10-2x) and the height would be x

a)
V = x(10-2x)(10-2x) , where 0 < x < 5

b) V = x(100 - 40x + 4x^2)
= 100x - 40x^2 + 4x^3
dV/dx = 100 - 80x + 12x^2
= 0 for a max of V
12x^2 - 80x + 100=0
3x^2 - 20x + 25 = 0
(x-5)(3x - 5) = 0
x = 5/3 or x = 5, but x ≠ 5 according to our domain

x = 5/3 will maximize the volume

Answer 2

i don't understand what is happening on the third line of B

Answer 3

I assumed you are studying Calculus, since this is a typical Calculus question.

Tell me otherwise

Answer 4

Im only in Precalc

Answer 5

can you just tell me how you got to dV/dx = 100-80x-12^2

Answer 6

I took the derivative of the function, which is one of the basic concepts in the Calculus course.

If you are not yet at that point, then this question should not have been put to you, since there is no other practical way to do this question.

I suppose you could use a "trial -and-error" method, that is, take different values of x in the above domain and see what kind of volume you would get.
e.g.
let x = 1
V = 1(8)(8) = 64
let x = 2
V = 2(6)(6) = 72 , which is more than 64
let x = 3
V = 3(4)(4) = 48 , ahh smaller
so lets try x = 2.5
V = 2.5(5)(5) = 62.5
how about 1.75
V = 1.75(6.5)(6.5) = 73.93 , heh, that's the biggest so far
how about 5/3 or 1.66666..
v = (5/3)(20/3)(20/3) = 74.074... ahh, even bigger
how about 1.65 , which is slighly < 5/3
v = 1.65(6.7)(6.7) = 74.0685 , a bit less
and finally how about x = 1.67 , a bit more than 5/3
V = 1.67(6.66)(6.66) = 74.0738.. , ahhh, a bit less than above

looks like my answer of x = 5/3 is right