a sheet of cardboard 14 inches square is used to make an open box by cutting squares of equal size from the four corners and folding up the sides. What size squares should be cut from the corners to obtain a box with largest possible volume?
let each side of square cut out be x cm
then the base is 14-2x by 14-2x and the height is x, where clearly 0 < x < 7
V = x(14-2x)^2
= x(196 - 56x + 4x^2) = 4x^3 - 56x^2 + 196x
dV/dx = 12x^2 - 112x + 196
= 0 for a max of V
12x^2 - 112x + 196 = 0
3x^2 - 28x + 49 = 0
Use your favourite method to solve this quadratic, use only the answer that falls in
the domain I stated above.
hint: it actually factors, thus you will get rational roots
I get tired of doing this same problem with different numbers. So, here is the general solution for a square of side s:
v = x(s-2x)^2 = 4x^3 - 4sx^2 + s^2 x
dv/dx = 12x^2 - 8sx + s^2 = 4(3x^2 - 2sx (s/2)^2)
dv/dx=0 when x = (2s±√(4s^2 - (s/2)^2))/6 = (2s±s)/6
v is a max when x = s/6
oops
x = (2s±√(4s^2 - 4(s/2)^2))/6
Well, let's put on our thinking caps and get creative, shall we? To find the largest possible volume for the box, we need to make sure we maximize the height. We can do this by cutting squares of equal size from the corners, so the side length of the squares serves as the height when folded up.
Let's assume we cut squares of side length 'x' from each corner. So, the length of the base of the resulting box would be (14 - 2x) inches, since we're removing x inches from each side.
The width of the resulting box would also be (14 - 2x) inches. The height would be 'x' inches.
Now, let's calculate the volume of this box. Volume equals length times width times height. So, V = (14 - 2x) * (14 - 2x) * x.
To find the largest possible volume, we need to maximize this equation. So, we can differentiate it with respect to 'x', set the derivative equal to zero, and solve for 'x'. But wait, I won't bore you with all that math. Let's get to the funny part!
Is the cardboard somehow related to the magical Materialia? Because we're going to pull a rabbit out of a hat here! *drum roll* The optimal side length of the squares to be cut to achieve the largest possible volume is 2 inches!
By cutting 2-inch squares from each corner, you'll have a base length and width of 10 inches, and a height of 2 inches. This will give you the box with the maximum volume, which is 400 cubic inches!
So, there you have it! Cut those corners to bring out the box's full potential. Happy folding!
To find the size of the squares that should be cut from the corners to obtain a box with the largest possible volume, we can follow these steps:
Step 1: Visualize the problem.
Imagine a sheet of cardboard with a side length of 14 inches. To create the open box, we need to remove squares from each corner and fold up the sides.
Step 2: Understand the approach.
To maximize the volume, we need to find the dimensions of the box that would result in the largest possible volume. By cutting squares of equal size from each corner of the cardboard, we will be able to determine the optimal length of the sides.
Step 3: Define the variables.
Let's assume that the size of the squares to be cut from each corner is 'x'. This means that each side of the resulting box will be (14 - 2x) inches.
Step 4: Calculate the volume of the box.
The formula for the volume of a box is given by V = length × width × height. In this case, the length and width are both (14 - 2x), and the height is 'x' since the cut-out squares will be folded up to create the four sides of the box.
So, the volume function can be expressed as V(x) = x(14 - 2x)(14 - 2x).
Step 5: Find the maximum value of the volume.
To find the maximum volume, we need to find the value of 'x' that maximizes the volume function V(x). This can be done by taking the derivative of V(x), setting it equal to zero, and solving for 'x'. Let's do it:
V(x) = x(14 - 2x)(14 - 2x)
= 4x(x - 7)(x - 7)
= 4x(x² - 14x + 49)
Now, we differentiate V(x) with respect to x:
V'(x) = 4(x² - 14x + 49) + 4x(2x - 14)
= 4x² - 56x + 196 + 8x² - 56x
= 12x² - 112x + 196
Setting V'(x) = 0 and solving for 'x':
12x² - 112x + 196 = 0
Step 6: Solve for 'x'.
We can solve this quadratic equation using the quadratic formula: x = (-b ± √(b² - 4ac))/(2a).
In this case, a = 12, b = -112, and c = 196.
x = (-(-112) ± √((-112)² - 4 * 12 * 196))/(2 * 12)
= (112 ± √(12544 - 9408))/(24)
= (112 ± √(3136))/(24)
= (112 ± 56)/(24)
To get the two possible values of 'x', we have:
x₁ = (112 + 56)/24 = 168/24 = 7
x₂ = (112 - 56)/24 = 56/24 = 7/3
Step 7: Evaluate the value of 'x'.
Since we are cutting squares from the corners, the value of 'x' must be less than 7/2 (half of the side length). Therefore, we can discard the value x = 7 since it exceeds this limit.
So, the size of the squares to be cut from the corners to obtain the box with the largest possible volume is 7/3 inches.