An open box having a square base is to be constructed from 32 square inches of material. What should be the dimensions of the box to obtain a maximum volume?
To solve this problem, we can use calculus to find the dimensions of the box that will yield the maximum volume.
Let's denote the dimensions of the square base as x and the height of the box as h.
Since the box is open, we have to consider that the total surface area will consist of the area of the square base and the four sides of the box.
The surface area of the square base is x^2, and since there are four sides of equal dimensions, the surface area of each side is xh. Therefore, the total surface area of the box is:
Surface Area = x^2 + 4xh
We are given that the total material available is 32 square inches. Therefore, we can set up the equation as:
x^2 + 4xh = 32
To find the maximum volume, we need to express the volume of the box in terms of a single variable, either x or h. Since we want to find the dimensions of the box, it is more convenient to express the volume in terms of x.
The volume of the box is given by:
Volume = x^2 * h
To eliminate h, we can express it in terms of x using the equation for the surface area:
h = (32 - x^2) / (4x)
Substituting this expression for h into the equation for volume, we get:
Volume = x^2 * (32 - x^2) / (4x)
Simplifying, we have:
Volume = (1/4) * x * (32 - x^2)
To find the dimensions that yield the maximum volume, we need to find the critical points of this function. We can do this by taking the derivative of the volume function and setting it equal to zero:
dV/dx = 0
We differentiate the volume function with respect to x:
dV/dx = (1/4) * (32 - x^2) - (1/4) * x * (2x)
Simplifying, we get:
dV/dx = (1/4) * (32 - 3x^2)
Setting this equal to zero, we have:
(1/4) * (32 - 3x^2) = 0
Solving for x, we get:
32 - 3x^2 = 0
3x^2 = 32
x^2 = 32/3
x = √(32/3)
x ≈ 2.83
Now that we have the value of x, we can substitute it back into the equation for h:
h = (32 - x^2) / (4x)
h = (32 - (2.83)^2) / (4 * 2.83)
h ≈ 2.83
Therefore, the dimensions of the box that yield the maximum volume are approximately:
Length = Width = x ≈ 2.83 inches
Height = h ≈ 2.83 inches
Note that these are approximate values rounded to two decimal places.