an open box with a square base is to be constructed from 48 square inches of material. What dimensions will produce a box with maximum volume?
The way I read your question:
You have 48 square inches available in total to make the box.
so let the base be x by x inches and the height y inches
total area available = x^2 + 4xy = 48
y = (48 - x^2)/(4x)
volume = x^2 y
= x^2 (48 - x^2)/(4x)
= 12x - x^3/4
d(volume)/dx = 12 - (3/4)x^2
= 0 for a max of volume
(3/4)x^2 = 12
x^2 = 16
x = √16 = 4
if x = 4, then y = (48 - 16)/16 = 2
the box should have a base of 4 inches by 4 inches and a height of 2 inches
To find the dimensions that will produce a box with the maximum volume, we can use the derivative of the volume function.
Let's denote the length of one side of the square base as "x" inches, and the height of the box as "h" inches.
The area of the square base is then x * x = x^2 square inches.
The area of each side of the box (excluding the base) is x * h square inches, and since the box is open, there are 4 sides in total, so the total area of the sides is 4 * x * h = 4xh square inches.
Therefore, the total surface area is given by:
x^2 + 4xh = 48 square inches.
We need to express the volume in terms of a single variable, either in terms of x or h. Since we want to find the dimensions, it's best to express the volume in terms of x:
The volume of the box is given by V = x^2 * h cubic inches.
Using the surface area equation, we can solve for h:
h = (48 - x^2) / (4x).
Now we can express the volume, V, in terms of x only:
V = x^2 * h = x^2 * [(48 - x^2) / (4x)] = (48x - x^3) / 4.
To find the maximum volume, we need to find the critical points of the volume function. This is done by finding where the derivative of V with respect to x is equal to zero.
Let's differentiate V with respect to x:
dV/dx = (48 - 3x^2) / 4.
Now we set the derivative equal to zero:
(48 - 3x^2) / 4 = 0.
Simplifying the equation:
48 - 3x^2 = 0.
Rearrange the equation:
3x^2 = 48.
Divide both sides by 3:
x^2 = 16.
Take the square root of both sides:
x = ±4.
Since we are dealing with dimensions, the length cannot be negative, so x = 4.
Now we can substitute x = 4 back into the surface area equation to find h:
h = (48 - x^2) / (4x) = (48 - 4^2) / (4 * 4) = (48 - 16) / 16 = 32 / 16 = 2.
Therefore, the dimensions that will produce a box with maximum volume are a square base with sides of length 4 inches and a height of 2 inches.
To determine the dimensions that will produce a box with maximum volume, we need to define the variables involved and set up an equation. Let's denote the side length of the square base as 'x' and the height of the box as 'h'.
1. Start with the given information: The box is constructed from 48 square inches of material. This means the total surface area of the box, which includes the base and the four sides, is 48 square inches.
2. Calculate the total surface area of the box:
- The area of the square base is x * x = x^2 square inches.
- The area of all four sides (height x width) is 4xh square inches.
- The total surface area is x^2 + 4xh square inches.
3. Set up the equation for the total surface area:
x^2 + 4xh = 48
4. Now, we need to express the volume in terms of one variable to find its maximum value. The volume of the box is given by:
Volume = base area * height
Volume = x^2 * h cubic inches
Volume = x^2h cubic inches
5. Rewrite the volume equation in terms of one variable (h):
Volume = 48/h
6. To find the dimensions for maximum volume, we need to maximize the volume function by finding the critical points. To do this, take the derivative of the volume equation with respect to 'h':
V'(h) = -48/h^2
7. Set V'(h) equal to zero to find the critical point:
-48/h^2 = 0
8. Solve for 'h':
h^2 = -48/0
h = 0 (disregard this solution because height can't be zero or negative)
9. Therefore, there are no critical points for the volume equation.
10. We also need to consider the boundary conditions. Since the material used is fixed at 48 square inches, there will be a constraint on the surface area of the box. We know that the surface area is given by:
x^2 + 4xh = 48
11. Re-arrange the equation to solve for x:
x = (48 - 4xh) / h
12. Substitute the expression for x in the volume equation:
Volume = ((48 - 4xh) / h)^2 * h
13. Simplify the equation:
Volume = (48 - 4xh)^2 / h
14. Now, we can use the surface area equation to eliminate one variable. Substitute the value of x from the surface area equation into the volume equation:
Volume = (48 - 4[(48 - 4xh) / h]h)^2 / h
15. Simplify the equation further:
Volume = ((48h - 4(48 - 4xh))^2) / h
16. Expand and simplify:
Volume = ((48h - 192 + 16xh)^2) / h
17. Now, we have the volume equation in terms of only one variable, 'h'. We can differentiate this expression with respect to 'h' to find the critical points.
18. Take the derivative of the volume equation with respect to 'h':
V'(h) = (2(48h - 192 + 16xh)(48 - 4xh) - (48h - 192 + 16xh)^2) / h^2
19. Set V'(h) equal to zero to find the critical points:
(2(48h - 192 + 16xh)(48 - 4xh) - (48h - 192 + 16xh)^2) / h^2 = 0
20. Solve the equation to find the values of 'h' that make the derivative zero.
21. Once you find the critical points, plug them into the volume equation and calculate the corresponding values of 'x' to determine the dimensions that will produce a box with maximum volume.
By following these steps, you can find the dimensions that will achieve maximum volume for the given amount of material.