assuming the cardboard is square, then if the cut has length x,
v = x(4-2x)^2
max volume occurs when dv/dx=0, at x = 2/3
A box with an open top is been constructed from a piece of cardboard that is 4m wide by cutting a out a square from each side of that corner and bending up the side . Find the maximum volume
2 answers
Let the side of each cut-out square be x m
base = 4-2x by 4-2x
height = x
volume = x(4-2x)^2 = 16x - 16x^2 + 4x^3
d(volume)/dx = 16 - 32x + 12x^2
=0 for a max of volume
divide each term by 4
3x^2 - 8x + 4 = 0
solve for x
Make sure you reject the inadmissible root, remember 0 < x < 2 or else the dimensions
make no sense
base = 4-2x by 4-2x
height = x
volume = x(4-2x)^2 = 16x - 16x^2 + 4x^3
d(volume)/dx = 16 - 32x + 12x^2
=0 for a max of volume
divide each term by 4
3x^2 - 8x + 4 = 0
solve for x
Make sure you reject the inadmissible root, remember 0 < x < 2 or else the dimensions
make no sense
1 answer