An open top bos with a square base is to be constructed from 192 sq ft of material. What should be the dimensions of the box if the volume is to be a maximum? What is the maximum volume?
base --- x by x ft
height -- y ft
area = 4xy + x^2
192 = 4xy + x^2
y = (192 - x^2)/(4x)
V = x^2 y
= x^2 (192-x^2)/(4x)
= 48x - (1/4)x^3
dV/dx = 48 - (3/4)x^2 = 0 for a max of V
(3/4)x^2 = 48
x^2 = 64
x = 8
then y = (192-64)/(32) = 4
max Volume = (8^2)(4) = 256
Oh, the dimensions of the box? Let me think outside the box for a moment, literally! To maximize the volume, we should give it some square-titude. Since it's an open-top box with a square base, let's assume the side length of the base is "x". So, the base area would be x^2, and we have 192 sq ft of material to work with.
Now, remember that the material won't just cover the base, but also the four sides. Each side would have an area of x times the height, which we'll call "h". Since there are four sides, the total area of the four sides would be 4xh.
But here's a side-splitting trick: the height "h" should be equal to the side length "x" itself to maximize the volume! This means the box is essentially a cube. So, we can rewrite the equation as:
x^2 + 4x^2 = 192
Simplifying further, we get:
5x^2 = 192
By solving for "x", we find x ≈ 6.93 ft.
And voila! The maximum volume of this wonderfully square-tastic box would be approximately (6.93 ft)^3, which is roughly 348.86 cubic feet.
So, now you can create your own box and bask in the glory of maximum volume!
To find the dimensions that maximize the volume of the box, we need to determine the side length of the square base and the height of the box. Let's proceed step-by-step:
Step 1: Let's assume the side length of the square base is 'x'. Since it is a square base, all sides are of equal length.
Step 2: The area of the square base is given by the formula: Side Length * Side Length = x * x = x^2.
Step 3: Since the box is open-top, the material required will be used to form the base and the sides of the box, excluding the top surface. So, the material used will be equal to the area of the base plus the area of the sides.
Step 4: Since the base is a square, the area of the base is x^2.
Step 5: The material used for the sides of the box forms a rectangle when unrolled. The length of each side of the rectangle will be equal to the height of the box. So, the total length of the material used for the sides is 4 * height.
Step 6: The total amount of material used is the sum of the material used for the base and sides, which is x^2 + 4 * height.
Step 7: According to the problem, the total material available is 192 sq ft. So, we can write the equation: x^2 + 4 * height = 192.
Step 8: We want to maximize the volume of the box, which is given by the formula: Volume = base area * height. The base area is x^2, so the volume is x^2 * height.
Step 9: We can now find an expression for height in terms of x and substitute it in the volume equation.
Step 10: Rearranging the equation from step 7, we have height = (192 - x^2) / 4.
Step 11: Substituting this expression for height in the volume equation, we get:
Volume = x^2 * ((192 - x^2) / 4) = (x^2 * (192 - x^2)) / 4.
Step 12: To find the maximum volume, we can differentiate the volume equation with respect to x, set the derivative equal to zero, and solve for x.
d(Volume)/dx = (192x - 3x^3) / 4.
Setting this equal to zero:
(192x - 3x^3) / 4 = 0.
192x - 3x^3 = 0.
3x(64 - x^2) = 0.
Using the zero-product property, we have two cases:
Case 1: 3x = 0, which gives x = 0. This is not a valid solution since the box would have zero dimensions.
Case 2: 64 - x^2 = 0.
x^2 = 64.
Taking the square root on both sides, we have:
x = ±8.
Since the sides of a square cannot be negative, we'll consider x = 8.
Step 13: Now that we have the value of x, we can substitute it back into the height equation from step 10 to find the value of height.
height = (192 - x^2) / 4 = (192 - 8^2) / 4 = (192 - 64) / 4 = 128 / 4 = 32.
So, the dimensions of the box that maximize the volume are a square base with side length 8 ft and a height of 32 ft.
Step 14: Finally, substituting these values into the volume equation, we find the maximum volume:
Volume = x^2 * height = 8^2 * 32 = 64 * 32 = 2048 cubic ft.
Therefore, the maximum volume of the box is 2048 cubic ft.
To determine the dimensions of the box that would maximize its volume, we need to set up an optimization problem. Let's proceed step by step:
Step 1: Identify the variables
Let's assume the side length of the square base is 'x' and the height of the box is 'h'. These will be the variables we need to find.
Step 2: Define the objective function
We want to maximize the volume of the box. The volume of a rectangular prism (box) is given by V = length × width × height. In this case, the length and width are both equal to 'x', so the volume can be expressed as V = x^2 × h.
Step 3: Express the constraints
The constraint given in the problem is that the material to construct the box has an area of 192 sq ft. The material used for construction includes the bases and the sides of the box. Since the base is a square, the total area of the bases is 2 × (x^2). The material for the sides of the box is the remaining area, which is equal to the sum of the four sides: 4 × (x × h). Therefore, we have the constraint: 2x^2 + 4xh = 192.
Step 4: Solve for one variable in terms of the other
From the constraint equation, we can solve for 'h' in terms of 'x':
2x^2 + 4xh = 192
h = (192 - 2x^2) / (4x)
h = (96 - x^2) / (2x)
Step 5: Substitute the expression for 'h' in the objective function
Substituting the expression for 'h' in the volume equation, we get:
V = x^2 × (96 - x^2) / (2x)
V = (48x - x^3) / 2
V = 24x - (x^3 / 2)
Step 6: Differentiate the volume equation
To find the maximum volume, we need to differentiate the volume equation with respect to 'x'. Taking the derivative:
dV/dx = 24 - (3x^2 / 2)
Step 7: Set the derivative equal to zero and solve for 'x'
To find the maximum, we set the derivative equal to zero and solve for 'x':
24 - (3x^2 / 2) = 0
48 - 3x^2 = 0
3x^2 = 48
x^2 = 16
x = ±√16
x = ±4
Since we are considering the dimensions of a box, we take the positive value: x = 4.
Step 8: Find the corresponding 'h' value
Substituting the value of 'x' into the constraint equation:
2x^2 + 4xh = 192
2(4^2) + 4(4)h = 192
32 + 16h = 192
16h = 160
h = 10
The dimensions that would maximize the volume of the box are a square base with side length of 4 ft and a height of 10 ft.
Step 9: Calculate the maximum volume
Substituting the values of 'x' and 'h' into the volume equation:
V = x^2 × h
V = 4^2 × 10
V = 160 cubic feet
Therefore, the maximum volume of the box is 160 cubic feet when the dimensions are a square base with side length of 4 feet and a height of 10 feet.