A closed box is to be a rectangular solid with a square base and volume of 12 ft^3. Find the most economical dimensions if the top of the box is twice as expensive as the sides and bottom.
Let's assume that the dimensions of the square base of the box are x by x, and the height of the box is h.
Since the box is a rectangular solid with a square base, the volume of the box is given by the formula V = length x width x height. In this case, we have a square base, so the length and width are the same and equal to x. Therefore, the volume equation becomes V = x^2 * h.
Given that the volume of the box is 12 ft^3, we can write the equation as:
x^2 * h = 12
Next, we need to find the surface area of the box to determine the cost. The surface area (SA) can be calculated by adding the areas of the six sides of the box. Since the square base has four sides, each with an area of x * h, and the other two sides have areas of x^2, the total surface area is given by:
SA = 4(x * h) + 2(x^2)
Based on the problem, the cost of the top of the box, which is one of the four sides with an area of x * h, is twice as expensive as the other sides, which are the other three sides and the bottom with areas of x * h. Let's assume the cost of the latter is C, then the cost of the top is 2C.
The cost equation can be written as:
Cost = C * (4(x * h) + 2(x^2)) + 2C * (x * h)
Since we want to find the most economical dimensions, we need to minimize the cost function. We can solve for x and h by substituting the volume equation (x^2 * h = 12) into the cost equation.
Substituting x^2 * h = 12 into the cost equation, we get:
Cost = C * (4(x * h) + 2(x^2)) + 2C * (x * h)
Cost = C * (4(x * h) + 2(x^2)) + 2C * (12 / h)
Now, let's take the derivative of the cost function with respect to h and set it to zero to find a critical point. Differentiating the cost function, we get:
dCost / dh = C * (4x - 24 / h^2) - 24C / h^2
Setting dCost / dh = 0, we can solve for h:
0 = C * (4x - 24 / h^2) - 24C / h^2
Simplifying the equation, we get:
4x - 24 / h^2 = 24 / h^2
4x = 48 / h^2
x = 12 / h^2
Now, substitute the value of x into the volume equation (x^2 * h = 12):
(12 / h^2)^2 * h = 12
144 / h^4 * h = 12
144 / h^3 = 12
144 = 12h^3
h^3 = 144 / 12
h^3 = 12
h = 2
Substituting h = 2 into x = 12 / h^2, we get:
x = 12 / (2^2)
x = 12 / 4
x = 3
Therefore, the most economical dimensions of the box are a square base with dimensions 3 ft by 3 ft and a height of 2 ft.
To find the most economical dimensions for the box, we can start by letting x be the length of one side of the square base, and y be the height of the box.
Since the volume of the box is given as 12 ft^3, we can write the equation: x^2 * y = 12.
To find the cost of the box, we need to consider that the cost of the top is twice as expensive as the sides and bottom. Let's call the cost of one square foot of the sides and bottom "c". Therefore, the cost of one square foot of the top is 2c.
The area of the sides and bottom of the box can be calculated as follows: 4x^2.
The area of the top of the box is: x^2.
Thus, the total cost of the box is: 4x^2 * c + x^2 * 2c = 4cx^2 + 2cx^2 = 6cx^2.
To find the most economical dimensions, we need to minimize the cost function 6cx^2 while satisfying the volume equation x^2 * y = 12.
To solve for the most economical dimensions, we can use optimization techniques such as calculus. We need to differentiate the cost function with respect to x and set it to zero.
d(6cx^2)/dx = 12cx = 0.
From this, we can see that c must be equal to zero or x must be equal to zero. However, x cannot be equal to zero since it represents the length of a side of the square base, and we are assuming the box has non-zero dimensions.
Therefore, c must be equal to zero. This means that the sides and bottom are free (cost zero) while the top has no area. Consequently, the box would have no lid, making its volume infinite.
In conclusion, there are no positive dimensions that would minimize the cost of the box, as creating a rectangular solid with a square base and volume of 12 ft^3 under the given constraints is not possible.