A rectangular box with a square base and top is to be made to contain 1250 cubic feet. The material for the base costs 35 cents per square foot, for the top 15 cents per square foot, and for the sides 20 cents per square foot. Find the dimensions that will minimize the cost of the box.
To find the dimensions that will minimize the cost of the box, we need to first define the variables involved. Let's call the length and width of the base of the box "x" and the height of the box "h".
Since the box has a square base, the length and width are equal, so we have x = x = x.
The volume of the box is given as 1250 cubic feet, so we can write the equation: x * x * h = 1250.
Now, let's find an expression for the cost of the materials used to construct the box.
The cost of the base is given as 35 cents per square foot, so the cost of the base is 35 * x * x.
The cost of the top is given as 15 cents per square foot, so the cost of the top is 15 * x * x.
The cost of the four sides is given as 20 cents per square foot, so the cost of the sides is 20 * x * h.
The total cost of the box can be expressed as: Cost = cost of base + cost of top + cost of sides = 35 * x * x + 15 * x * x + 20 * x * h.
Now, we need to express the height "h" in terms of "x" to simplify the expression further.
From the volume equation, h = 1250 / (x * x).
Substituting this value of "h" into the cost equation, we get: Cost = 35 * x * x + 15 * x * x + 20 * x * (1250 / (x * x)).
Simplifying this expression further, we get: Cost = 50 * x * x + 25000 / x.
To minimize the cost, we need to find the value of "x" that minimizes the cost. To do so, we can take the derivative of the cost equation with respect to "x" and set it equal to zero, and then solve for "x".
dCost/dx = 100 * x - 25000 / (x^2) = 0.
Multiplying through by x^2, we get: 100 * x^3 - 25000 = 0.
Solving for "x", we find: x^3 = 250, x = (250)^(1/3) ≈ 6.299.
So, the approximate dimensions that will minimize the cost of the box are:
Length = Width = x = (250)^(1/3) ≈ 6.299 ft,
Height = h = 1250 / (x * x) ≈ 1250 / (6.299 * 6.299) ≈ 31.571 ft.
To minimize the cost of the box, we need to find the dimensions that will minimize the surface area. Let's assume the length and width of the base of the box are x and x, respectively, and the height is h.
1. Volume of the box:
Since the box has a square base, the volume is given by:
V = x * x * h = x^2 * h = 1250
2. Surface area of the box:
The surface area is the sum of the areas of the base, top, and four sides:
S = 2x^2 + 4xh
3. Express h in terms of x:
From the volume equation, we can solve for h:
h = 1250 / (x^2)
4. Substitute h in the surface area equation:
S = 2x^2 + 4x * (1250 / (x^2)) = 2x^2 + 5000 / x
5. Find the derivative of S with respect to x:
dS/dx = 4x - 5000 / x^2
6. Set the derivative equal to zero and solve for x:
4x - 5000 / x^2 = 0
4x^3 -5000 = 0
x^3 = 1250
x = cube root of 1250 ≈ 10.85
7. Substitute x back into the volume equation to find h:
h = 1250 / (10.85)^2 ≈ 10.85
Therefore, the dimensions that will minimize the cost of the box are approximately:
Length = Width (x) ≈ 10.85 feet
Height (h) ≈ 10.85 feet
You would assume the side of the square (base or top) to be x (feet).
The height can be expressed in terms of x and the volume. (V=L*W*H=x²*H)
You would then express the sum of the costs, C, made up of
1. the cost of the base = area of base * unit cost of base,
2. the cost of the sides and
3. the cost of the top.
Now differentiate C(x) with respect to x and equate C'(x)=0. Solve for x.