The material for the base of a box will cost three times as much as the material for the sides and top of the box. The box must have a volume of 200m cubed. Find the most efficient way to built this box.
There is additional information needed.
I am guessing from previous questions like this that the base is a square.
Let the base be x metres by x metres, an let the height be y metres.
then (x^2)y = 200 and y = 200/x^2
Cost = 3(x^2) + 1(4xy + x^2)
= 4x^2 + 4xy
= 4x^2 + 4x(200/x^2)
= 4x^2 + 800/x
Cost' = 8x - 800/x^2 = 0 for a max/min of Cost
8x = 800/x^2
x^3 = 100
x = 100^(1/3) = 4.64 m
then y = 200/4.64^2 = 9.28 m
the box should have a base of 4.64m by 4.64m and a height of 9.28 m
I also have a feeling that the volume was given as 2000, then the answer would have been exactly x=10 and y = 20
which would be in the same height:base length of 2:1
To find the most efficient way to build this box, we need to minimize the cost of the materials while meeting the given requirements. Let's break down the problem step by step:
Step 1: Define variables
Let's assume the cost of the material for the sides and top of the box is represented by "x" (in some currency), and the cost of the material for the base is three times that, so it would be "3x".
Step 2: Determine the dimensions of the box
To find the most efficient way to build the box, we need to determine the dimensions of the box that will minimize the cost while maintaining the volume requirement.
Let's assume the length, width, and height of the rectangular box are represented by "l", "w", and "h" respectively.
Given that the volume of the box is 200 m^3, we can write the equation:
l * w * h = 200
Step 3: Express one variable in terms of the others
Since the cost of the base material is three times that of the sides and top, we can express the cost of the base material in terms of "x" (the cost of the sides and top).
The cost of the sides and top of the box = x
The cost of the base of the box = 3x
Step 4: Express the cost function
The cost function, C, can be expressed as:
C = 3x + x = 4x
Step 5: Express one variable in terms of the others
To solve for one variable in terms of the others, we can rearrange the volume equation:
l * w * h = 200
h = 200 / (l * w)
Step 6: Substitute the expression for one variable into the cost function
Substituting the expression for "h" in the cost function, we have:
C = 4x = 4x(l * w) / 200
Step 7: Determine the critical points
To find the critical points where the cost function is minimum, we can take the derivative of the cost function with respect to either "l" or "w", and set it equal to zero.
dC/dl = 0 or dC/dw = 0
Taking the derivative of the cost function (using the quotient rule), we have:
dC/dl = 4x * (w * (d/dl)(200) - 200 * (d/dl)(w)) / (200^2)
Simplifying, we get:
dC/dl = 4x * (w * (-200)/ (200^2))
dC/dl = -4xw / 200
Setting dC/dl = 0, we have:
-4xw / 200 = 0
Solving for "w", we get "w = 0".
Similarly, taking the derivative with respect to "w" and setting it equal to zero, we would find that "l = 0".
Step 8: Evaluate the critical points
Since "w" and "l" cannot be zero (as they represent dimensions), we can conclude that there are no critical points to evaluate.
Step 9: Consider physical constraints
Since we have no critical points to evaluate, we need to consider the physical constraints to determine a feasible solution.
The dimensions of the box cannot be negative or zero, and since the volume is given as 200 m^3, the dimensions need to be positive.
Step 10: Determine the feasible solution
Given the physical constraints, we need to find positive values for the length, width, and height that satisfy the volume requirement.
We can explore various combinations of positive dimensions that satisfy the volume equation, such as:
- l = 10 m, w = 4 m, h = 5 m (since 10 * 4 * 5 = 200)
- l = 5 m, w = 8 m, h = 5 m (since 5 * 8 * 5 = 200)
Note: There are multiple combinations of positive dimensions that satisfy the volume requirement, and each combination will yield a different cost. The most efficient way to build the box would be to find the combination that minimizes the cost, given the dimensions.
In summary, there are multiple feasible solutions for building the box that meets the volume requirement. To determine the most efficient method, consider the cost of the materials based on the given cost function and dimensions. The combination of dimensions that yields the minimum cost would be the most efficient way to build the box.