A rectangular box is to have a square base and a volume of 20 ft3. If the material for the base costs 37¢/square foot, the material for the sides costs 10¢/square foot, and the material for the top costs 13¢/square foot, determine the dimensions of the box that can be constructed at minimum cost.
x = ft
y = ft
area of top = x^2
area of sides = 4 x y
area of top = x^2
volume = x^2 y = 20
cost of bottom = 37 x^2
cost of top = 13 x^2
cost of sides = 10 * 4 * x y = 40 x y
total cost = c = 50 x^2 + 40 x y
volume constraint: y = 20/x^2
so
c = 50 x^2 + 40*20/x
dc/dx = 0 for max or min
0 = 100 x - 800/x^2
x = 8/x^2
x^3 = 8
x = 2
then
y = 20/4 = 5
To determine the dimensions of the box that can be constructed at minimum cost, we need to minimize the cost function. The cost function is the sum of the cost of the base, sides, and top of the box.
Let's start by determining the dimensions of the box.
Given that the box has a square base, we can assume that the length, width, and height of the box are all equal. Let's represent this common side length as "x".
The volume of the box is given as 20 ft^3. Since the base is square, the area of the base is x^2, and the height of the box is also x. Therefore, we have the equation:
Volume = x^2 * x = 20 ft^3
Simplifying this equation, we get:
x^3 = 20
Now, to minimize the cost function, we need to express the cost as a function of x.
The cost of the base is given as 37¢/square foot, and since the area of the base is x^2, the cost of the base is:
Cost of base = 37 * x^2 ¢
The cost of the sides is given as 10¢/square foot, and since there are 4 sides (excluding the base), the cost of the sides is:
Cost of sides = 4 * 10 * x^2 ¢
The cost of the top is given as 13¢/square foot, and since the area of the top is x^2, the cost of the top is:
Cost of top = 13 * x^2 ¢
Now, we can express the total cost function as the sum of the costs:
Total cost = Cost of base + Cost of sides + Cost of top
= 37 * x^2 + 4 * 10 * x^2 + 13 * x^2
Simplifying further, we get:
Total cost = 37x^2 + 40x^2 + 13x^2
= 90x^2
To minimize the cost function, we can take the derivative of the total cost function with respect to x and set it equal to zero:
d(Total cost) / dx = 180x = 0
Solving for x, we get:
x = 0
Since we cannot have a box with zero dimensions, we need to consider the endpoints of the feasible range of x.
The feasible range of x is restricted by the volume equation: x^3 = 20
Taking the cube root of both sides, we get:
x = 20^(1/3)
Thus, the value of x that gives the minimum cost is approximately 2.714 ft (rounded to three decimal places).
Now that we have the value of x, we can find the corresponding values of y (width) and z (height) by substituting x into the volume equation:
x * x * x = 20
(20^(1/3)) * (20^(1/3)) * (20^(1/3)) = 20
Therefore, the dimensions of the box that can be constructed at minimum cost are approximately:
x = 2.714 ft
y = 2.714 ft
z = 2.714 ft