We want to build a rectangular cardboard box with no top and of given volume V
using the least quantity of cardboard. Find the length of the box's edges by �fnding the minimum
of the total surface function. Verify that the point you found is indeed a minimum applying the
second derivatives test on such function.
To find the length of the box's edges that will minimize the quantity of cardboard used, we can approach the problem by using calculus.
Let's start by defining the variables we will use:
- Length of the box: L
- Width of the box: W
- Height of the box: H
We are given that the volume of the box is V. Therefore, we have the equation:
V = L * W * H
To find the minimum quantity of cardboard used, we need to minimize the total surface area of the box. The total surface area consists of the area of the four sides of the box and the area of the bottom.
The area of the four sides is given by:
A_sides = 2LH + 2WH
The area of the bottom is given by:
A_bottom = LW
The total surface area can be expressed as:
A_total = A_sides + A_bottom
= 2LH + 2WH + LW
We want to minimize A_total, subject to the constraint V = LWH.
Now, let's express A_total in terms of a single variable. We can rewrite it using the volume constraint:
A_total = 2LH + 2WH + LW
= 2LH + 2W(V/LW) + LW
= 2LH + 2V/W + LW
Since V and W are constants, we can treat them as such while taking the derivative with respect to L.
To find the minimum, we need to find the critical points by taking the derivative of A_total with respect to L and setting it equal to zero:
dA_total/dL = 2H + V/W = 0
Simplifying the equation, we get:
2H = -V/W
H = -V/2W
Since H cannot be negative, we can disregard the negative solution. Therefore, we have the height of the box as H = V/2W.
Substituting this value of H back into the volume equation, we get:
V = L * W * (V/2W)
V = L * V/2
L = 2
Thus, the length of the box's edges that will minimize the quantity of cardboard used is L = 2.
To verify that this point is indeed a minimum, we can apply the second derivative test. We need to find the second derivative of A_total with respect to L:
d^2A_total/dL^2 = 0
Since the second derivative is zero, we cannot make a conclusion regarding the second derivative test. Therefore, we need to examine the behavior of A_total near L = 2 to determine if it is a minimum or not.
By analyzing the behavior of A_total for values of L slightly greater and slightly smaller than 2, we can confirm if 2 is the minimum point.
Keep in mind that this solution assumes that only one box of length 2 can be built with the given volume V. If there are other possible dimensions that can achieve the same volume, they should also be considered and compared for the least amount of cardboard used.