a closed box with square base is to have a V= 2000in^3. The material for the top and bottom of the box is to cost $3 per square inch and the material for the sides is the cost $1.50 per square inch. If the cost of the material is to be the least, find the dimensions
To find the dimensions of the closed box that minimize the cost, we can start by expressing the volume of the box in terms of its dimensions.
Let's assume the base of the box is a square with side length "x" inches, and the height of the box is "h" inches.
Given that the volume of the box is V = 2000 in^3, we have:
V = x^2 * h
We need to minimize the cost, so let's find the expression for the cost in terms of "x" and "h".
The cost of the top and bottom is $3 per square inch, and since there are two identical square bases, the cost of the top and bottom combined is:
Cost of top and bottom = 2 * (x^2 * 3)
The cost of the sides is $1.50 per square inch, and there are four sides of the box, each with an area of "x * h". Hence, the cost of the sides is:
Cost of sides = 4 * (x * h * 1.5)
Therefore, the total cost can be represented as:
Total cost = Cost of top and bottom + Cost of sides
= 2 * (x^2 * 3) + 4 * (x * h * 1.5)
We want to minimize the total cost, so let's differentiate it with respect to "x" and "h" and set the resulting equations to zero:
d(Total cost)/dx = 0
d(Total cost)/dh = 0
Differentiating the total cost with respect to "x" gives:
d(Total cost)/dx = 0
6x + 6h = 0
x + h = 0 ---- (Equation 1)
Differentiating the total cost with respect to "h" gives:
d(Total cost)/dh = 0
4x + 6h = 0 ---- (Equation 2)
Now, we have a system of equations (Equation 1 and Equation 2) to solve for "x" and "h".
To solve these equations, we can multiply Equation 1 by 6 and Equation 2 by -6 to eliminate the "h" term:
6(x + h) = 0
-6(4x + 6h) = 0
Simplifying these equations gives:
6x + 6h = 0
-24x - 36h = 0
Now, let's add these two equations together:
6x + 6h - 24x - 36h = 0
-18x - 30h = 0
Dividing by -6 gives:
3x + 5h = 0 ---- (Equation 3)
Now, we can substitute Equation 1 into Equation 3 to solve for "x":
x + h = 0 (Equation 1)
=> h = -x
Substituting h = -x into Equation 3 gives:
3x + 5(-x) = 0
3x - 5x = 0
-2x = 0
x = 0
Since the dimensions cannot have a length of 0, this indicates that there is no minimum cost for a box with a square base and volume of 2000 in^3.
Hence, it is not possible to minimize the cost for the given volume of 2000 cubic inches.
To find the dimensions of the closed box with a square base that minimize the cost of materials, we need to optimize two things simultaneously—the surface area and the volume.
Let's call the length of each side of the square base 'x', and the height of the box 'h'.
Since the base is square, its area is x^2 square inches. The surface area of the four sides of the box is 4xh square inches.
The volume of the box is given as V = 2000 cubic inches, which means x^2 * h = 2000.
We are trying to minimize the cost, which is based on the surface area. The cost of the top and bottom covers is $3 per square inch, and the cost of the four side covers is $1.50 per square inch. So, the total cost is T = 3(2x^2) + 1.5(4xh) = 6x^2 + 6xh.
Now, let's solve this problem using the method of Lagrange multipliers, which helps optimize a function given a constraint.
First, we set up the equations:
1. The volume constraint: x^2 * h = 2000
2. The cost function: T = 6x^2 + 6xh
Next, we define the Lagrangian function (L) as:
L(x, h, λ) = T - λ(V - 2000)
Differentiating the Lagrangian with respect to x, h, and λ, we have:
∂L/∂x = 12x + 6h - λ(2x^2)
∂L/∂h = 6x - λ(x^2)
∂L/∂λ = 2000 - x^2 * h
Setting these partial derivatives to zero, we solve the resulting system of equations to find the critical points.
Solving the equations, we find:
12x + 6h = 2λx^2 → (1)
6x = λ(x^2) → (2)
2000 = x^2 * h → (3)
From equation (2), we have two possible scenarios:
Scenario 1: λ = 6 and x ≠ 0
Scenario 2: x = 0 (which corresponds to a degenerate case we can ignore for this problem)
In Scenario 1, we substitute λ = 6 into equation (1) and divide through by x to get:
12 + 6h/x = 2x → 6h/x = 2x - 12 → 6h = 2x^2 - 12x → h = (1/3)x^2 - 2x
Substituting h = (1/3)x^2 - 2x in equation (3), we have:
2000 = x^2 * [(1/3)x^2 - 2x]
2000 = (1/3)x^4 - 2x^3
Multiplying through by 3 to eliminate the fraction:
6000 = x^4 - 6x^3
Rearranging:
x^4 - 6x^3 - 6000 = 0
This is a quartic equation that can be solved numerically, either using a graphing calculator or computer software.
Once we find the positive real solution(s) for x, we substitute them back into h = (1/3)x^2 - 2x to find the corresponding values of h. These values of x and h will give us the dimensions of the box that minimize the cost of materials.
Note that this method finds the exact dimensions that minimize the cost, but numerical approximations are often used to simplify the process.