A mailing service places a limit of 36 inches on the combined length and girth of (distance around) a package to be sent parcel post. What dimensions of a rectangular box with square cross-section will contain the largest volume that can be mailed. (Hint: There are two different girths.)
Let's denote the side length of the square cross-section as x and the height of the rectangular box as h.
The combined length and girth of the box is given by the formula:
2x + 2h = 36
Solving for h in terms of x, we get:
h = 18 - x
The volume of the box is given by:
V = x^2 * h
Substitute h = 18 - x into the equation for V:
V = x^2 * (18 - x)
V = 18x^2 - x^3
To find the maximum volume, we take the derivative of V with respect to x and set it equal to 0:
dV/dx = 36x - 3x^2
36x - 3x^2 = 0
3x(12 - x) = 0
x = 0 or x = 12
Since x cannot be 0 (it is a dimension of a box), the only possible value for x is x = 12. Now we can find the corresponding value of h:
h = 18 - x
h = 18 - 12
h = 6
Therefore, the dimensions of the rectangular box with square cross-section that will contain the largest volume that can be mailed are:
Width = 12 inches
Height = 6 inches