A box with a square base and no top is to be built with a volume of
6912 in^3. Find the dimensions of the box that requires the least amount of material. How much material is required at the minimum?
To find the dimensions of the box that requires the least amount of material, we can use calculus and optimization techniques. Let's denote the side length of the square base as x and the height of the box as h.
First, we need to express the amount of material required in terms of x and h. The box has a square base and no top, so the surface area of the box can be calculated as follows:
Surface Area = base area + 4(side area)
The base area is simply x^2, and the side area can be calculated by multiplying the perimeter of the base (4x) by the height (h). Therefore, the surface area can be expressed as:
Surface Area = x^2 + 4xh
Next, we know that the volume of the box is 6912 in^3, so we have the equation:
Volume = base area * height
6912 = x^2 * h
h = 6912 / x^2
Substituting this value of h into the surface area equation, we get:
Surface Area = x^2 + 4x(6912 / x^2)
Surface Area = x^2 + 27648 / x
To find the dimensions that minimize the surface area, we can take the derivative of the surface area with respect to x and set it equal to zero:
d(Surface Area)/dx = 2x - 27648/x^2 = 0
Solving this equation, we find x = 24. Taking the second derivative, we get:
d^2(Surface Area)/dx^2 = 2 + 55296/x^3
Since the second derivative is positive, we can conclude that x=24 gives us the minimum value of the surface area.
To find the corresponding height, we can substitute x=24 into the equation for h:
h = 6912 / (24^2) = 12
Therefore, the dimensions of the box that requires the least amount of material are:
- Base side length: 24 inches
- Height: 12 inches
To calculate the minimum amount of material required, substitute x=24 and h=12 into the surface area equation:
Surface Area = (24^2) + 4(24)(12) = 576 + 1152 = 1728 in^2
So, the minimum amount of material required is 1728 square inches.
To find the dimensions of the box that requires the least amount of material, we can use the concept of optimization.
Let's assume that the length of the sides of the square base is x inches, and the height of the box is h inches.
The volume of the box is given by the formula V = x^2 * h. Given that V = 6912 in^3, we can write the equation as x^2 * h = 6912.
Now, we need to find the dimensions (x and h) that minimize the amount of surface area of the box, as the surface area represents the amount of material required.
The surface area of the box can be broken down into three parts: the sides of the square base (4 * x^2), the base itself (x^2), and the height multiplied by the perimeter of the square base (4 * x * h).
So the total surface area S is given by S = 4x^2 + x^2 + 4xh.
To minimize S, let's take the derivative of S with respect to x and h, set them equal to zero, and solve the resulting equations to find the values of x and h.
dS/dx = 8x + 4h = 0 (equation 1)
dS/dh = 4x = 0 (equation 2)
From equation 2, we find that x = 0.
Substituting x = 0 into equation 1, we get 4h = 0, which implies h = 0.
However, having dimensions of zero does not make sense in this context, so we need to consider the limits of this problem.
Since the volume V is given to be 6912 in^3, and V = x^2 * h, we can rewrite equation 1 as 4h = 6912 / x^2.
Substituting this value of 4h into equation 1, we get 8x + (6912 / x^2) = 0.
Multiplying both sides by x^2, we get 8x^3 + 6912 = 0.
We can now solve this cubic equation to find the value of x.
Once we have the value of x, we can substitute it back into the equation 1 to find the value of h.
Once we have the values of x and h, we can calculate the total surface area S using the formula S = 4x^2 + x^2 + 4xh.
The dimensions (x and h) that minimize the surface area of the box are the required dimensions for the box that requires the least amount of material.
To find the actual amount of material required at the minimum, we can substitute the values of x and h into the surface area formula and calculate it.
Surface area: 2Lh+2Wh+LW
volume=Lhw or h= 6912/Lw
surface area=2L(6912)/Lw+2w*6912/Lw + Lw
= 2*6912/w+2*6912/L+Lw
now, it has a square base (I just saw that, so L=W)
=2*6912/x+2*6912/x+x^2
take the derivative of area /dx
set to zero.
0=-4*6912/x^2+2x
or x^3=8(6912)
L=W=38.0976 inch
solve for h now h=6912/Lw