A box (with no lid) is to be constructed from a sheet of card board by cutting the squares from corners and folding up the sides. Suppose the original sheet of card board measures 16 inches by 16 inches. What would the size of the squares removed to maximize the volume of the resulting box?
v = x (16 - 2x)²
solve for the volume
take the 1st derivative (dv/dx), and set equal to zero
solve for x
test the two solutions
To find the size of the squares to be removed, we need to maximize the volume of the resulting box. Let's assume that we remove squares with side length 'x' from each corner.
The length of the resulting box would be (16 - 2x) inches, and the width would also be (16 - 2x) inches. The height of the box would be 'x' inches.
So, the volume V of the box can be found by multiplying the length, width, and height:
V = (16 - 2x) * (16 - 2x) * x
Now, we can simplify the equation and find the value of 'x' that maximizes the volume. We can start by expanding the equation:
V = (256 - 64x + 4x^2) * x
V = 256x - 64x^2 + 4x^3
To find the maximum volume, we can take the derivative of V with respect to 'x' and set it equal to zero. Let's find the derivative:
dV/dx = 256 - 128x + 12x^2
Setting dV/dx equal to zero and solving for 'x':
0 = 256 - 128x + 12x^2
128x = 256 + 12x^2
12x^2 - 128x + 256 = 0
Now, we can solve this quadratic equation using the quadratic formula:
x = (-(-128) ± √((-128)^2 - 4(12)(256))) / (2(12))
x = (128 ± √(16384 - 12288)) / 24
x = (128 ± √4096) / 24
x = (128 ± 64) / 24
So, we have two possible values for 'x':
1. x = (128 + 64) / 24 = 192 / 24 = 8 inches
2. x = (128 - 64) / 24 = 64 / 24 = 2.67 inches (approximately)
Since we cannot have a square with a fractional length, we can discard the second solution. Therefore, the size of the squares to be removed to maximize the volume of the resulting box is 8 inches.