What is the maximum volume in cubic inches of an open box to be made from a 16 in by 30 in 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.
length = 30 - 2x
width = 16 - 2x
V = (30-2x)(16-2x) = 4 (15-x)(8-x)
dV/4dx = (15-x)(-1) + (8-x)(-1)
= 2x^2 -23
fo min or max dV/dx = 0
x^2 = 23/2
x = sqrt(23/2)
did bottom area, not volume !
V = 4 x (15-x)(8-x)
V/4 = x(120 - 23 x + x^2)
V/4 = 120 x -23 x^2 + x^3
dV/4dx = 120 -46 x + 3 x^2 = 0 for min or max
solve quadratic
https://www.mathsisfun.com/quadratic-equation-solver.html
I get x = 12 or x = 3.3333333.... or 10/3
12 does not work. 16 - 2*12 is a very small side
So try 10/3
V = (10/3)(16 - 10/3)(30 - 10/3)
To find the maximum volume of the open box, we need to determine the dimensions that will maximize the volume. Let's start by analyzing the problem.
We have a rectangular piece of cardboard with dimensions 16 inches by 30 inches. We need to cut squares of equal sides from the four corners and fold up the remaining sides to create an open box.
Let's assume we cut squares with side length x from each corner. After cutting, the dimensions of the remaining cardboard will be (16 - 2x) inches by (30 - 2x) inches. The height of the box will be x inches.
To find the maximum volume, we need to maximize the function that represents the volume of the box.
Volume of the box = length * width * height
Volume = (16 - 2x) * (30 - 2x) * x
To maximize this volume, we can find the derivative of the volume function with respect to x and set it equal to zero.
Let's differentiate the volume function:
dV/dx = (16 - 2x) * (30 - 2x) * 1 + (16 - 2x) * x * (-2) + (30 - 2x) * x * (-2)
Simplifying, we get:
dV/dx = -4x^2 + 88x - 480
To find the critical points, we set the derivative equal to zero and solve for x:
-4x^2 + 88x - 480 = 0
To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, let's use factoring:
-4x^2 + 88x - 480 = 0
x^2 - 22x + 120 = 0
(x - 10)(x - 12) = 0
From this, we get two possible solutions: x = 10 inches and x = 12 inches.
Next, we need to determine which value of x provides the maximum volume. We can do this by evaluating the second derivative of the volume function at these critical points.
To find the second derivative, we differentiate the first derivative:
d^2V/dx^2 = -8x + 88
Now, let's evaluate the second derivative at the critical points:
d^2V/dx^2 (x = 10) = -8(10) + 88 = 8
d^2V/dx^2 (x = 12) = -8(12) + 88 = -16
Since the second derivative at x = 10 is positive, the volume function has a local minimum at x = 10. Similarly, since the second derivative at x = 12 is negative, the volume function has a local maximum at x = 12.
Therefore, x = 12 represents the value that gives us the maximum volume.
To find the maximum volume, substitute this value of x back into the volume function:
Volume = (16 - 2(12)) * (30 - 2(12)) * 12
Volume = 4 * 6 * 12
Volume = 288 cubic inches
So, the maximum volume of the open box is 288 cubic inches.