A rectangular box with a square bottom and a volume of 256 cubic feet is to be constructed. The top and bottom cost $ .10 per square foot to make and the four sides cost $ .05 per square foot to make. Find the approximate dimensions of the box which would minimize its cost.
let each side of the base be x ft
let the height of the box be y ft
V= (x^2)(y) = 256
y = 256/x^2
form the cost equation ....
C = 2(.10)x^2 + 4(.05)xy
= .2x^2 + 2x(256/x^2)
= .2x^2 + 512/x
C' = .4x - 512/x^2 = 0 for min of C
I get x^3 = 1280
x = 10.86 and y = 2.17
check my arithmetic
To find the approximate dimensions that would minimize the cost of the box, we need to determine the dimensions of the square bottom and the height of the box.
Let's assume the side length of the square bottom is x, and the height of the box is h.
The volume of a rectangular box is given by the formula V = Length * Width * Height.
Given that the volume of the box is 256 cubic feet, we can write:
x * x * h = 256
Simplifying this equation, we have:
x^2 * h = 256 ---(Equation 1)
The cost of the top and bottom is $0.10 per square foot, while the cost of the four sides is $0.05 per square foot.
The cost of the top and bottom can be calculated as 2 * Area of the square bottom * $0.10, and the cost of the four sides can be calculated as 4 * Area of the square bottom * $0.05.
The area of the square bottom is x * x = x^2.
Therefore, the cost of the top and bottom is 2 * x^2 * $0.10 = $0.20 * x^2, and the cost of the four sides is 4 * x^2 * $0.05 = $0.20 * x^2.
Therefore, the total cost C(x) is given by:
C(x) = $0.20 * x^2 + $0.20 * x^2 = $0.40 * x^2 ---(Equation 2)
To minimize the cost, we can take the derivative of C(x) with respect to x and set it equal to 0 to find the critical points.
dC/dx = 0.40 * 2x = 0
Simplifying this equation, we have:
0.8x = 0
Dividing both sides by 0.8, we get:
x = 0
However, x = 0 does not make sense in the context of our problem. Therefore, there are no critical points.
Let's find the dimensions of the box by substituting x from equation 1 into equation 2.
C(x) = $0.40 * x^2 = $0.40 * (256 / h)
To minimize the cost C(x), we need to minimize the function C(x) = $0.40 * (256 / h).
Since h > 0, we can conclude that C(x) is minimized when h is maximum.
From equation 1, we have x^2 * h = 256. Solving for h, we get:
h = 256 / x^2
Substituting this value of h into C(x), we obtain:
C(x) = $0.40 * (256 / (256 / x^2))
C(x) = $0.40 * x^2
Therefore, the cost is only dependent on x^2, and it does not depend on the value of h.
To minimize the cost, we need to minimize x^2.
Since x is the side length of the square bottom, it must be positive.
Therefore, the approximate dimensions that would minimize the cost of the box are x = 16 and h = 256 / x^2 = 256 / (16^2) = 256/256 = 1.
Thus, the approximate dimensions of the box that would minimize its cost are:
- The side length of the square bottom is 16 feet.
- The height of the box is 1 foot.
To find the approximate dimensions of the box that minimize its cost, we need to express the cost function in terms of a single variable and then find the critical points using calculus.
Let's denote the length of each side of the square bottom as 'x' and the height as 'h'. Since the box is rectangular, the other two sides of the base will also have lengths 'x'.
The volume of the box is given as 256 cubic feet, so we can write an equation:
Volume = x * x * h = 256
To minimize the cost, we need to determine the dimensions that minimize the cost function. Let's denote the cost function as 'C'. The top and bottom have an area of x * x = x^2, and the four sides have an area of 2x * h.
The total cost of the box can be calculated as:
C = 0.10 * (2 * x^2) + 0.05 * (2x * h)
Simplifying this, we get:
C = 0.20x^2 + 0.10xh
Now, we have two variables, 'x' and 'h', in the cost function. To deal with this, we need to express 'h' in terms of 'x' using the volume equation.
From the volume equation, we have:
h = 256 / (x^2)
Substituting this value of 'h' into the cost function, we get:
C = 0.20x^2 + 0.10x * (256 / (x^2))
Simplifying this further, we have:
C = 0.20x^2 + 25.6 / x
Now, to find the minimum of this cost function, we need to take the derivative with respect to 'x', set it equal to zero, and solve for 'x'.
dC/dx = 0.40x - 25.6 / (x^2)
Setting dC/dx = 0, we have:
0.40x - 25.6 / (x^2) = 0
Multiply both sides by (x^2) to get rid of the denominator:
0.40x^3 - 25.6 = 0
Simplifying further:
0.40x^3 = 25.6
Divide both sides by 0.40 to isolate 'x':
x^3 = 64
Taking the cube root of both sides:
x = ∛64 = 4
Now that we've found 'x', we can substitute it back into the volume equation to find the value of 'h':
h = 256 / (x^2) = 256 / (4^2) = 256 / 16 = 16
Therefore, the approximate dimensions of the box that minimize its cost are a square bottom with side length 4 feet and a height of 16 feet.