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A square sheet of cardboard with a side 16 inches is used to make an open box by cutting squares of equal size from the four co...Asked by Anonymous
a sheet of cardboard 14 inches square is used to make an open box by cutting squares of equal size from the four corners and folding up the sides. What size squares should be cut from the corners to obtain a box with largest possible volume?
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Answered by
Reiny
let each side of square cut out be x cm
then the base is 14-2x by 14-2x and the height is x, where clearly 0 < x < 7
V = x(14-2x)^2
= x(196 - 56x + 4x^2) = 4x^3 - 56x^2 + 196x
dV/dx = 12x^2 - 112x + 196
= 0 for a max of V
12x^2 - 112x + 196 = 0
3x^2 - 28x + 49 = 0
Use your favourite method to solve this quadratic, use only the answer that falls in
the domain I stated above.
hint: it actually factors, thus you will get rational roots
then the base is 14-2x by 14-2x and the height is x, where clearly 0 < x < 7
V = x(14-2x)^2
= x(196 - 56x + 4x^2) = 4x^3 - 56x^2 + 196x
dV/dx = 12x^2 - 112x + 196
= 0 for a max of V
12x^2 - 112x + 196 = 0
3x^2 - 28x + 49 = 0
Use your favourite method to solve this quadratic, use only the answer that falls in
the domain I stated above.
hint: it actually factors, thus you will get rational roots
Answered by
oobleck
I get tired of doing this same problem with different numbers. So, here is the general solution for a square of side s:
v = x(s-2x)^2 = 4x^3 - 4sx^2 + s^2 x
dv/dx = 12x^2 - 8sx + s^2 = 4(3x^2 - 2sx (s/2)^2)
dv/dx=0 when x = (2s±√(4s^2 - (s/2)^2))/6 = (2s±s)/6
v is a max when x = s/6
v = x(s-2x)^2 = 4x^3 - 4sx^2 + s^2 x
dv/dx = 12x^2 - 8sx + s^2 = 4(3x^2 - 2sx (s/2)^2)
dv/dx=0 when x = (2s±√(4s^2 - (s/2)^2))/6 = (2s±s)/6
v is a max when x = s/6
Answered by
oobleck
oops
x = (2s±√(4s^2 - 4(s/2)^2))/6
x = (2s±√(4s^2 - 4(s/2)^2))/6
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