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Tom makes an open box from a rectangular piece of metal by cutting equal squares from the four corners and turning up the sides. The piece of metal measures 60 cm by 100 cm. What are the dimensions of the box with the maximum volume?

2007-06-22 14:34:31 · 4 answers · asked by slow_math 1 in Science & Mathematics Mathematics

4 answers

since there are 2 dimensions, these would be the length and height of the box.

For max volume, pick the length of the surface as the bigger dimension (it will be squared to get the volume)

therefore volume = (length)^2 * height
= 100^2 * 60
= 600000cm2

2007-06-22 14:39:51 · answer #1 · answered by mathnerd 2 · 0 2

If the unknown length of the sides of the equal squares is x, then the dimensions of the box are:
L = 100 - 2x (an x on either end)
W = 60 - 2x
H = x
Volume of a box = L*W*H, so,
V = (100 - 2x)(60 - 2x)x
= (6000 - 320x + 4x^2)x
= 6000x - 320x^2 + 4x^3
Now to find the maximum volume, take the derivative, set it equal to zero, and solve for x.
V' = 6000 - 640x + 12x^2 = 0
Dividing out a 4 gives:
4(3x^2 - 160x + 1500) = 0
3x^2 - 160x + 1500 = 0
Using the quadratic formula, you get:
x = [160 +/- sqrt((-160)^2 - 4*3*1500)]/2*3
= [160 +/- sqrt(25600 - 18000)]/6
= [160 +/- sqrt(7600)]/6
= [160 + 87.178]/6 or [160 - 87.178]/6
= 41.196 or 12.137
Now to check which gives a max.
You can see that if you plug in x = 41.196 into (60 - 2x), you get a negative number, which would give you a negative width. So, that value is no good. To verify that x = 12.137 gives a max, check the value of V' around x = 12.137:
V'(12) = a positive number, V'(13) = a negative number. Thus, based on the first derivative test, x = 12.137 is a relative maximum. Thus, the dimensions of the box with maximum volume are:
L = 100 - 2x = 100 - 2(12.137) = 75.726 cm
W = 60 - 2x = 60 - 2(12.137) = 35.726 cm
H = 12.137 cm
Hope this helps.
PS. there's a slight error in Steve's factoring out the 4.

2007-06-22 22:08:21 · answer #2 · answered by Lee 3 · 2 0

Let side of square that's cut out = s
V = l*w*h
L = 60 - 2s
W = 100 - 2s
H = s
V = (60-2s)(100-2s)s
V = 6000s - 320s^2 + 4s^3
V' = 6000 - 640s + 12s^2
0 = 6000 - 640s + 12s^2
0 = 2000 - 160s + 3s^2
0 = (3s - 100)(s - 20)
s = 33 1/3, 20
If you cut 33 1/3 out of both end of the 60cm side, you get negative, so s = 20
Dimensions 20x20x60

2007-06-22 22:04:31 · answer #3 · answered by Steve A 7 · 0 0

a=length
b=breath
c=height
v=volume

v=abc
b+2c=60
a+2c=100
--->a-b=40
--->a=40+b

--->v=(40+b)bc
--->dV/dc=40b+b^2
--->dV/db=40c+2bc
40b+b^2=40c+2bc for maximum volume,
so substitute b+2c=60 to solve it.

2007-06-22 22:38:02 · answer #4 · answered by Anonymous · 0 0

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