A rectangular box with a volume of 320 cubic units is to be constructed with a square base and top. The cost per square foot of the bottom is 15 cents, for the top 10 cents, and for the sides is 2.5 cents. Find the dimensions that minimize the cost of constructing the box
let the base be x by x
let the height be y
x^ y = volume
x^2 y = 320
y = 320/x^2
cost = 15x^2 + 10x^2 + 4(2.5)xy
= 25x^2 + 10x(320/x^2)
= 25x^2 + 3200/x
d(cost)/dx = 50x - 3200/x^2 = 0 for a min of cost
50x = 3200/x^2
x^3 = 64
x = 4
then y = 320/16 = 20
the box should have a base of 4 by 4 and a height of 20 units
To find the dimensions that minimize the cost of constructing the box, we need to set up an optimization problem using calculus.
Let's assume the length of the sides of the square base and top of the box is x. Since the volume of the box is given as 320 cubic units, we can set up the equation:
x^2 * h = 320
where h represents the height of the box.
Next, we need to calculate the cost of constructing the box. The cost of the bottom and top will be proportional to their areas, and the cost of the sides will be proportional to their surface area.
The cost of the bottom = 0.15 * x^2
The cost of the top = 0.10 * x^2
The cost of the sides = 4 * 0.025 * x * h = 0.1 * x * h
Now, we can express the total cost C in terms of x and h:
C = (0.15 * x^2) + (0.10 * x^2) + (0.1 * x * h)
To minimize the cost, we need to minimize this function with respect to x and h. To do this, we can take partial derivatives of C with respect to x and h, and set them equal to zero:
∂C/∂x = 0.3 * x + 0.1 * h = 0
∂C/∂h = 0.1 * x = 0
From the second equation, we see that x = 0.
Substituting x = 0 into the first equation, we get:
0.3 * x + 0.1 * h = 0
0.1 * h = 0
So, h = 0.
However, this doesn't make sense in the context of constructing a box. A box with zero height or length is not a box.
Thus, it seems there is a mistake or contradiction in the problem, as there is no feasible solution for the given conditions. Please double-check the given information or context.
To find the dimensions that minimize the cost of constructing the box, we need to determine the dimensions of the square base.
Let's assume the side length of the square base is "x" units. Since the box has a square base and a square top, the dimensions of the box will be x by x by h, where h represents the height of the box.
The volume of a rectangular box is given by the formula: Volume = length × width × height. In this case, the volume is given as 320 cubic units, so we have:
320 = x × x × h
320 = x^2h
Now, let's express h in terms of x:
h = 320 / (x^2)
Next, we need to determine the cost of constructing the box. We are given the cost per square foot for the bottom, top, and sides. The cost for the bottom is $0.15 per square foot, the top is $0.10 per square foot, and the sides are $0.025 per square foot.
The cost of constructing the bottom is given by: Cost_bottom = area_bottom × cost_per_square_foot_bottom
Since the bottom is a square with side length "x", the area of the bottom is x^2 square units. Therefore, the cost of constructing the bottom is:
Cost_bottom = x^2 × $0.15
The cost of constructing the top is the same as the cost of constructing the bottom:
Cost_top = x^2 × $0.10
The cost of constructing the four sides is given by: Cost_sides = area_sides × cost_per_square_foot_sides
Since there are four sides, and each side has an area of x × h:
Cost_sides = 4 × (x × h) × $0.025
Cost_sides = 4 × x × (320 / (x^2)) × $0.025
The total cost of constructing the box is the sum of the costs for the bottom, top, and sides:
Total_cost = Cost_bottom + Cost_top + Cost_sides
Now, let's substitute the expressions we derived earlier for the costs:
Total_cost = (x^2 × $0.15) + (x^2 × $0.10) + (4 × x × (320 / (x^2)) × $0.025)
Simplifying the expression, we have:
Total_cost = (0.15x^2) + (0.10x^2) + (0.025 × 4 × 320 / x)
Total_cost = 0.25x^2 + (0.025 × 4 × 320 / x)
To minimize the cost, we need to find the minimum of this Total_cost function. To do that, we can take the derivative with respect to x, set it equal to 0, and solve for x.
d(Total_cost) / dx = 0.5x - (0.025 × 4 × 320 / x^2) = 0
Simplifying further, we have:
0.5x - (0.025 × 4 × 320 / x^2) = 0
0.5x = (0.025 × 4 × 320 / x^2)
0.5x^3 = 0.025 × 4 × 320
x^3 = (0.025 × 4 × 320) / 0.5
x^3 = 200
To solve for x, we cube root both sides:
x = ∛200
Approximating the value of ∛200, we find:
x ≈ 6.299
Therefore, the side length of the square base of the box that minimizes the cost is approximately 6.299 units.
To find the height, we can substitute this value of x back into the equation for h:
h = 320 / (x^2)
h = 320 / (6.299^2)
h ≈ 8.096
So, the approximate dimensions that minimize the cost of constructing the box are a square base with side length 6.299 units and a height of 8.096 units.