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
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
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
To find the maximum volume of the box, we need to determine the dimensions that would optimize it. Let's assume that the side length of the square cut out from each corner is "x".
If we cut out a square with side length "x" from each corner, the resulting dimensions of the cardboard would be:
Width = 4m - 2x
Length = 4m - 2x
Height = x
Now, the volume of the box can be calculated by multiplying these dimensions together:
Volume = (4m - 2x) * (4m - 2x) * x
To find the maximum volume, we need to find the value of "x" that will maximize this volume equation. We can do this by taking the derivative of the volume equation with respect to "x" and setting it equal to zero:
dV/dx = 0
Let's differentiate the volume equation with respect to "x":
dV/dx = (4m - 2x) * (4m - 2x) + 2x * (4m - 2x) - 2x * (4m - 2x) = 0
Simplifying this equation:
16m^2 - 16mx + 4x^2 + 8mx - 4x^2 = 0
16m^2 - 8mx = 0
Dividing through by 8:
2m^2 - mx = 0
Factoring out an "x":
x(2m - x) = 0
Setting each factor equal to zero and solving for "x":
x = 0 or 2m - x = 0
If we consider the dimensions of the cardboard (4m wide and 4m long), it's clear that the value of "x" cannot be zero, otherwise, there would be no cardboard left to form the sides of the box.
So, considering the second part of the equation:
2m - x = 0
x = 2m
Therefore, the optimal value for "x" to maximize the volume is half of the width of the cardboard.
Plugging this value back into the volume equation:
Volume = (4m - 2(2m)) * (4m - 2(2m)) * (2m)
Volume = (4m - 4m) * (4m - 4m) * (2m)
Volume = 0 * 0 * (2m)
Volume = 0
Hence, the maximum volume is zero. This implies that for the given dimensions of the cardboard, it is not possible to create a box with a maximum volume.