What is the maximum volume in cubic inches of an open box to be made from a 12-inch by 16-inch piece of cardboard by cutting out squares of equal sides from the four corners and bending up the sides? Your work must include a statement of the function and its derivative. Give one decimal place in your final answer.
v = (12-2x)(16-2x)x
dv/dx = 4(3x^2-28x+48)
set dv/dx=0 and solve for x.
To find the maximum volume of the open box, we need to determine the dimensions of the box that will maximize the volume.
Let's denote the length of the side that is cut out from each corner as x. Then, the dimensions of the box can be expressed as follows:
Length = (16 - 2x) inches
Width = (12 - 2x) inches
Height = x inches
The volume of the box can be calculated by multiplying these dimensions:
V(x) = (16 - 2x)(12 - 2x)(x) cubic inches
To find the maximum volume, we need to find the critical points of the volume function. We can do this by taking the derivative of V(x) and setting it equal to zero:
V'(x) = -4x^2 + 56x -192 = 0
Now, let's solve for x:
-4x^2 + 56x - 192 = 0
Dividing through by -4:
x^2 - 14x + 48 = 0
Factoring this quadratic equation:
(x - 6)(x - 8) = 0
Setting each factor equal to zero:
x - 6 = 0 or x - 8 = 0
So, the critical points are x = 6 and x = 8.
To determine if these critical points correspond to a maximum or a minimum, we can use the second derivative test. Taking the second derivative of V(x):
V''(x) = -8x + 56
Evaluating V''(6):
V''(6) = -8(6) + 56 = 8
Since the second derivative is positive for x = 6, it indicates a local minimum.
Evaluating V''(8):
V''(8) = -8(8) + 56 = -8
Since the second derivative is negative for x = 8, it indicates a local maximum.
Since we are interested in finding the maximum volume, we know that x = 8 corresponds to the maximum volume.
Substituting x = 8 into the volume function, V(x):
V(8) = (16 - 2(8))(12 - 2(8))(8) = 128 cubic inches
Therefore, the maximum volume of the open box is 128 cubic inches.
To find the maximum volume of the open box, we need to find the dimensions that would maximize the volume. Let's denote the side length of the squares to be cut from the corners as "x".
First, let's determine the dimensions of the box after cutting the squares from the corners.
The length of the box will be (16 - 2x) inches, as we remove two squares of sides "x" from the total length of 16 inches.
Similarly, the width of the box will be (12 - 2x) inches.
Now, we can calculate the volume of the box as the product of its length, width, and height. The height of the box, in this case, will be "x" inches.
Therefore, the volume of the box can be represented with the function V(x) = x(16 - 2x)(12 - 2x).
To find the maximum volume, we need to find the critical points of the function V(x) by taking its derivative and setting it equal to zero.
Let's find the derivative of V(x) with respect to x:
V'(x) = [16 - 2x][(12 - 2x)x' + x(12 - 2x)' - 2x(16 - 2x)']
= [16 - 2x][(12 - 2x) - 2x]
= (16 - 2x)(12 - 2x - 2x)
Simplifying the expression further:
V'(x) = (16 - 2x)(12 - 4x)
Setting V'(x) equal to zero and solving for x:
(16 - 2x)(12 - 4x) = 0
From this equation, we can find two possible values for x: x = 8 and x = 4.
To determine which value of x gives us the maximum volume, we need to compare the values of V(x) at these critical points. We also need to evaluate the endpoints of the interval, which are x = 0 (no squares cut) and x = 6 (cutting squares of side 6 from each corner would result in a negative volume, which is not possible).
Now, let's calculate the values of V(x) for the critical points and endpoints:
V(0) = 0(16)(12) = 0
V(4) = 4(16 - 2(4))(12 - 2(4)) = 4(8)(4) = 128
V(8) = 8(16 - 2(8))(12 - 2(8)) = 8(0)(-4) = 0
V(6) = 6(16 - 2(6))(12 - 2(6)) = 6(4)(0) = 0
From the calculations, we can see that the maximum volume occurs when x = 4 inches, with a volume of 128 cubic inches.
Therefore, the maximum volume of the open box is 128 cubic inches.