A rectangular box with a square base and cover is to have a volume of 2500 cubic feet. If the cost per square foot for the bottom is $2, for the top is $3, and for the sides is $1, what should the dimensions be in order to minimize the cost?
If the base has side x, and the height is h, then
x^2h = 2500
h = 2500/x^2
So, the cost to make the box is
c = 2(x^2) + 3(x^2) + 1*4*xh
= 5x^2 + 10000/x
dc/dx = 10x-10000/x^2
dc/dx=0 when x=10
so, the box is 10x10x25 ft
To minimize the cost, we need to find the dimensions of the rectangular box that satisfy the given volume condition while minimizing the cost per square foot.
Let's assume that the dimensions of the square base are x and x. The height of the box will then be h.
From the given information, we know that the volume of the rectangular box is 2500 cubic feet.
The volume of the rectangular box is given by:
Volume = Base area * Height
Since the base is square and has dimensions x by x, the base area is x * x = x^2. Therefore, the volume equation can be rewritten as:
2500 = x^2 * h
To minimize the cost, we need to minimize the surface area of the box. The surface area is the sum of the areas of the bottom, top, and four sides.
The cost for the bottom is $2 per square foot, so the cost of the bottom is 2 * x^2.
The cost for the top is $3 per square foot, so the cost of the top is 3 * x^2.
The cost for the sides is $1 per square foot, so the cost of the sides is 4 * x * h.
The total cost is the sum of the costs for the bottom, top, and sides:
Cost = 2 * x^2 + 3 * x^2 + 4 * x * h
Now, we need to find the dimensions that minimize the cost. To do this, we can use the given volume equation to express one variable in terms of the other and substitute into the cost equation.
From the volume equation, we have: h = 2500 / (x^2)
Substituting this into the cost equation:
Cost = 2 * x^2 + 3 * x^2 + 4 * x * (2500 / (x^2))
= 2 * x^2 + 3 * x^2 + (10000 / x)
Simplifying:
Cost = 5 * x^2 + (10000 / x)
To minimize the cost, we take the derivative of the cost equation with respect to x and set it equal to zero. We can then solve for x to find the critical points.
dCost/dx = 10 * x - (10000 / x^2) = 0
Multiplying through by x^2:
10 * x^3 - 10000 = 0
Simplifying:
x^3 - 1000 = 0
By factoring, we find that x = 10.
Now, we need to find the corresponding value of h. Substituting x = 10 into the volume equation:
2500 = 10^2 * h
2500 = 100 * h
h = 25
Therefore, the dimensions of the box that satisfy the given volume condition and minimize the cost are x = 10, x = 10, and h = 25.
Thus, the dimensions of the box should be 10 feet by 10 feet by 25 feet in order to minimize the cost.
To minimize the cost, we need to set up a function that represents the cost in terms of the dimensions of the rectangular box.
Let's assume the length, width, and height of the box are l, w, and h respectively.
We are given that the volume of the box is 2500 cubic feet, so we have the equation:
l * w * h = 2500
To find the cost, we need to consider the costs of the bottom, top, and sides.
The cost of the bottom is $2 per square foot, so the cost of the bottom is:
Cost of bottom = 2 * (l * w)
The cost of the top is $3 per square foot, so the cost of the top is:
Cost of top = 3 * (l * w)
The cost of the sides is $1 per square foot, so the cost of the sides is:
Cost of sides = 1 * (2lw + 2lh + 2wh) = 2lw + 2lh + 2wh
The total cost is the sum of the cost of the bottom, top, and sides:
Total cost = Cost of bottom + Cost of top + Cost of sides
Total cost = 2lw + 3lw + 2lh + 3lh + 2wh
Total cost = 5lw + 5lh + 2wh
To minimize the cost, we can differentiate the total cost with respect to one of the variables (l, w, or h) and set it to zero.
Let's differentiate the total cost with respect to l:
d(Total cost) / dl = 5w + 5h = 0
Simplifying, we have:
w + h = 0
Since we have two variables and only one equation, we can't directly solve for l, w, and h.
However, we are given that the box has a square base. This means l = w.
Substituting l = w into the equation w + h = 0, we get:
w + h = 0
Since the dimensions of a rectangular box cannot be negative, we can conclude that the dimensions that minimize the cost are l = w = 0.
However, this is not a valid solution.
Therefore, it appears that there may be a mistake in the problem statement or the given constraints.