A rectangular box has a square cross-section and the sum of its length and the perimeter of this cross-section is 2 m. If the length of the box is x m, show that its volume V m^3 is given by V = [x(2-x)^2]/16.
Hence find the maximum volume of the box.
2/27m^3
x=2/3
To solve this problem, we need to use the given information and algebraic manipulation to express the volume of the box in terms of its length (x).
Let's start by considering the cross-section of the box. Since it is square, each side of the square cross-section has a length of x/4 (since the perimeter is equal to 2m and the length of the box is x). Therefore, the area of the cross-section is (x/4)^2 = x^2/16.
Now, we need to find the height of the box. The sum of the length and perimeter of the cross-section is given as 2m. Since the length is x, the perimeter is 2(x/4) = x/2. Therefore, we can write the equation:
x + x/2 = 2
Simplifying this equation, we get:
2x + x = 4
3x = 4
x = 4/3
Now that we know the value of x, we can substitute it back into the expression for the area of the cross-section:
Area = (4/3)^2/16
Area = 16/9/16
Area = 1/9
Finally, we can use the formula for the volume of a rectangular box:
V = length * width * height
Plugging in the values, we have:
V = (4/3) * (1/9) * (x/4)
V = (4/27) * x/4
V = x/27
To find the maximum volume of the box, we need to find the maximum value of V. Since V = x/27, we can see that the maximum value occurs when x is the maximum value. From the given information, the length of the box can be at most 2m (as the sum of length and perimeter cannot exceed 2m).
Thus, the maximum volume is:
V = (2/27)^3 = 8/729 m^3