A box with a rectangular base and no top is to be made from a 9 inch by 12 inch piece of cardboard by cutting squares out of the corners and folding up the sides. What size (side-length) squares should be cut out to make the box have as much volume as possible?
If the squares have side x, then the volume is
v = x(9-2x)(12-2x) = 4x^3 - 42x^2 + 108x
dv/dx = 12x^2 - 84x + 108 = 12(x^2-7x+9)
now just find where dv/dx=0
To maximize the volume of the box, we need to find the dimensions that will give us the greatest amount of space. Let's start by visualizing the box.
The cardboard has dimensions of 9 inches by 12 inches. Let's assume that each side square that will be cut out has a side length of x inches.
We know that the length of the box will be 12 inches, and when we fold the sides up, the width will be decreased by 2x inches (since we cut out squares from both ends). Similarly, the height of the box will be decreased by 2x inches as well.
So, the length of the box will remain 12 inches, and the width and height will be (9 - 2x) inches.
The volume of the box can be calculated by multiplying the length, width, and height together:
Volume = length * width * height
= 12 * (9 - 2x) * (9 - 2x)
= 12 * (81 - 18x - 18x + 4x^2)
= 12 * (4x^2 - 36x + 81)
= 48x^2 - 432x + 972
To find the value of x that maximizes the volume, we need to find the maximum point of the quadratic equation. We can do this by finding the x-coordinate of the vertex.
The formula for finding the x-coordinate of the vertex of a quadratic equation in the form of ax^2 + bx + c is:
x = -b / (2 * a)
In our equation, a = 48, b = -432, and c = 972.
x = -(-432) / (2 * 48)
= 432 / 96
= 4.5
So, to maximize the volume, we should cut squares with side length 4.5 inches.
Please note that cutting squares of 4.5 inches from each corner may not be practical since it would result in removing more than the available cardboard area. In such cases, a possible solution would be to cut squares with a side length of 4 inches, which will provide a good approximation to maximize the volume while using the available cardboard area efficiently.
To determine the size of the squares that should be cut out to maximize the volume of the box, we need to understand the relationship between the dimensions of the cardboard and the resulting box.
Let's start by visualizing the process:
1. Cut squares of side length 'x' from each corner of the rectangular base of the cardboard.
2. Fold up the remaining flaps to form the sides of the box.
The resulting box would have the following dimensions:
- Length: 12 - 2x (since two squares of side length 'x' are removed)
- Width: 9 - 2x
- Height: x (the length of the squares cut out)
To find the volume, we multiply these dimensions together:
Volume = Length * Width * Height
= (12 - 2x) * (9 - 2x) * x
To maximize the volume, we need to find the derivative of this expression with respect to 'x' and set it equal to zero:
d/dx (Volume) = 0
Let's calculate the derivative:
d/dx [(12 - 2x) * (9 - 2x) * x] = 0
Expanding and simplifying the expression, we get:
-4x^3 + 42x^2 - 108x + 108 = 0
Unfortunately, this is a cubic equation, and finding the exact solution would be complicated. However, we can use numerical methods or estimations to find the approximate value of 'x' that maximizes the volume.
Using a graphing calculator, software, or online tools, we can plot the function y = -4x^3 + 42x^2 - 108x + 108 and find the maximum point.