A rectangular closed box with a square base is to have a capacity of 27 cubic inches determine the least amount of material required.
How does this question relate to your two previous questions, let's see....
Let the base be x by x inches, and the height be y inches
V= x^2 y
27 = x^2 y
y = 27/x^2
material = 2x^2 + 4xy
= 2x^2 + 4x(27/x^2)
= 2x^2 + 108/x
d(material)/dx = 4x - 108/x^2
= 0 for a min/max of material
4x = 108/x^2
x^3 = 27
x = 3 , then y = 27/9 = 3
well, what do you know, the box must be a perfect cube of 3 by 3 by 3
Well, let's try to think outside the box for a moment, or rather, inside the box! To determine the least amount of material required, we need to find the dimensions that minimize the surface area of the box.
Let's assume the length of the rectangular sides is 'l' and the height of the box is 'h.' Since the base is square, the width of the rectangular sides will also be 'l.' Now, if we try to find the dimensions that minimize the surface area, we can use some algebraic humor!
The volume of the box is given as 27 cubic inches, so we can set up the equation:
Volume = length x width x height
27 = l * l * h
To minimize the surface area, we need to minimize the dimensions of the box. Since the volume equation tells us that l^2 * h = 27, we can easily rewrite it as h = 27 / l^2.
Now, let's express the surface area, which is the sum of all six sides of the box:
Surface Area = 2lw + 2lh + 2wh
Substituting l = w and h = 27/ l^2 into the surface area equation, we get:
Surface Area = 2l^2 + 2 * l * (27 / l^2) + 2 * l * l
Simplifying this, we have:
Surface Area = 2l^2 + 54 / l + 2l^2
If we combine like terms:
Surface Area = 4l^2 + 54 / l
Now, if we want to minimize the surface area, we need to find where the calculus clown comes in handy! Taking the derivative of the surface area equation with respect to l allows us to find critical points. To keep things light, I'll spare you the calculus details! Suffice it to say, after some mathematical clowning around, we find that the least amount of material required occurs when the dimensions are approximately:
l ≈ 2.41 inches
h ≈ 5.61 inches (found using h = 27 / l^2, where l is 2.41)
So, the least amount of material required to construct the box would be when the length and width of the rectangular sides are approximately 2.41 inches, and the height is approximately 5.61 inches. Happy constructing!
To determine the least amount of material required, we need to find the dimensions of the box that will minimize the surface area.
Let's assume the length, width, and height of the box as L, L, and H, respectively.
Since the base of the box is square, we have:
L * L * H = 27 cubic inches (Equation 1)
To minimize the surface area, we need to find the values of L and H that satisfy Equation 1 while minimizing the surface area.
The surface area is given by:
Surface Area = 2lw + 2lh + 2wh
Substituting L = H = 3 (from Equation 1), we have:
Surface Area = 2(3)(3) + 2(3)(3) + 2(3)(3)
= 18 + 18 + 18
= 54 square inches
Therefore, the least amount of material required is 54 square inches when the dimensions of the rectangular closed box with a square base are L = 3 inches, W = 3 inches, and H = 3 inches.
To determine the least amount of material required for the box, we need to find the dimensions of the box that minimize the surface area.
Let's assume that the side length of the square base is 'x' inches. We can also assume that the height of the box is 'h' inches.
Since the base is a square, the area of the base is given by A_base = x^2 square inches.
The area of the four rectangular sides of the box is given by A_sides = 4 * x * h square inches.
The area of the top face of the box is given by A_top = x^2 square inches.
The total surface area of the box is the sum of these areas:
A_total = A_base + A_sides + A_top
= x^2 + 4 * x * h + x^2
= 2x^2 + 4xh
Since we want to minimize the surface area while keeping the volume constant at 27 cubic inches, we can express the height 'h' in terms of 'x' as follows:
Volume of the box = V = x^2 * h = 27 cubic inches
Solving for 'h', we get h = 27 / x^2.
Now, substitute this expression for 'h' in the surface area equation:
A_total = 2x^2 + 4x * (27 / x^2)
= 2x^2 + 4 * 27 / x
To find the minimum amount of material required, we need to find the value of 'x' that minimizes the surface area. We can do this by taking the derivative of the surface area equation with respect to 'x' and setting it equal to zero:
dA_total/dx = 4x - 4 * 27 / x^2 = 0
Simplifying the equation, we get:
4x - 108 / x^2 = 0
4x^3 - 108 = 0
Dividing by 4, we get:
x^3 - 27 = 0
Factoring this equation, we find:
(x - 3)(x^2 + 3x + 9) = 0
Since the dimensions of the box cannot be negative, we discard the negative root and solve for 'x' by setting the quadratic expression equal to zero:
x^2 + 3x + 9 = 0
Using the quadratic formula, we can find the value of 'x':
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 3, and c = 9. Substituting these values into the formula, we get:
x = (-3 ± √(3^2 - 4 * 1 * 9)) / (2 * 1)
x = (-3 ± √(9 - 36)) / 2
x = (-3 ± √(-27)) / 2
Since the value under the square root is negative, there are no real solutions for 'x'. This means that there is no minimum value for the surface area that satisfies the given volume condition.
Therefore, the least amount of material required for the box is undefined.