What is the maximum volume in cubic inches of an open box to be made from a 16-inch by 30-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.
L = 30 - 2 h
w = 16 - 2 h
V = L * w * h
V = (30-2h)(16-2h)h
take dv / dh and set it to zero
use only resulting h values that do not give you negative w when you subtract 2 h from 16
To find the maximum volume of the open box, we first need to determine the function that represents the volume of the box in terms of the length of the side of the squares being cut out.
Let's denote the length of the side of the square being cut out as 'x.' When the squares are cut out and the sides are folded up, the length of the box will be equal to the original length minus 2 times the length of the side of the square (30 - 2x), and the width will be equal to the original width minus 2 times the length of the side of the square (16 - 2x). The height of the box will be equal to the length of the side of the square (x).
Therefore, the volume V(x) of the box is given by the product of the length, width, and height:
V(x) = (30 - 2x)(16 - 2x)x
To find the maximum volume, we need to find the value of x that maximizes this function. We can do this by taking the derivative of V(x) with respect to x and setting it equal to zero.
Let's find the derivative of V(x):
V'(x) = 2x(16 - 2x) + (30 - 2x)(2) - 2x(30 - 2x)
Simplifying the equation:
V'(x) = 2x(16 - 2x) + 60 - 4x - 2x(30 - 2x)
V'(x) = 32x - 4x^2 + 60 - 4x - 60 + 4x^2
V'(x) = 28x
Setting V'(x) equal to zero:
28x = 0
x = 0
We have found that x = 0 is a critical point of the function V(x). However, since x represents the length of a side of a square, it cannot be equal to zero.
Thus, we need to check the endpoints of the feasible region which is the interval (0, 8) (since the original length is 16 and we can at most cut out squares with a side length of 8).
Evaluating the function V(x) at the endpoints:
V(0) = (30 - 2(0))(16 - 2(0))(0) = 0
V(8) = (30 - 2(8))(16 - 2(8))(8) = 448
Since the volume of the box cannot be negative, the maximum volume is V(8) = 448 cubic inches.
Therefore, the maximum volume of the open box is 448 cubic inches.