A box with a square-base is to be constructed with no top. The bottom face will cost 4 times as much as the sides. The volume of the box will be 0.5 m3. What dimensions (length, width and height) will minimize the cost to construct this box?
If the base has side length x and the box has height h, you know that
x^2 h = 1/2
So, h = 1/(2x^2)
The cost of the box will be
c = 4x^2 + 4(xh) = 4x^2 + 4x/(2x^2) = 4x^2 + 2/x
the minimum cost occurs when dc/dx = 0
c' = 8x - 2/x^2 = 2(4x^3-1)/x^2
c'=0 when x = 1/∛4
So the box with least cost is 1/∛4 x 1/∛4 x ∛2
or approximately 0.63 x 0.63 x 1.26
To minimize the cost of constructing the box, we need to find the dimensions (length, width, and height) that satisfy the given conditions while minimizing the cost.
Let's start by assigning variables to the dimensions:
- Let's say the length of the square base is denoted by 'x'.
- The height of the box will also be denoted by 'x' since it has a square base.
- The width of the box will be denoted by 'y'.
Now let's calculate the cost:
- The cost of the bottom face is 4 times the cost of each side, so the cost of the bottom face is 4y^2.
- The cost of each side is y times the height, which is y^2.
- The total cost is the cost of the bottom face plus the cost of the four sides: 4y^2 + 4y^2 = 8y^2.
We also know the volume of the box, which is given as 0.5 m^3:
- The volume of a rectangular box is length times width times height.
- In this case, the volume is x * y * x = x^2 * y = 0.5.
Now, we have two equations:
1. x^2 * y = 0.5 (equation for volume)
2. cost = 8y^2 (equation for cost)
To minimize the cost, we need to find the values of x and y that satisfy these equations.
To solve the equations, we can use substitution:
- From the volume equation, we can express x^2 in terms of y: x^2 = 0.5/y.
- Substituting this value of x^2 into the cost equation: cost = 8y^2 = 8 * (0.5/y) = 4/y.
To further simplify the equation, let's multiply both sides by y:
- y * cost = 4.
Now, we can solve for y:
- y = 4 / cost.
Since we want to minimize the cost, we can see that as the cost increases, the value of y decreases. Hence, as the cost approaches infinity, y approaches zero.
Therefore, in order to minimize the cost while maintaining the volume of 0.5 m^3, the dimensions should be:
- x = sqrt(0.5/y), where y is a very large number approaching infinity.
- y = 0, as it approaches zero.
In summary, to minimize the cost, the box should have a square base with one side approaching zero length, and the width should be a very large number.