A box with a square base and no top is to have a volume of 32 ft3. What dimensions use the least amount of material (in other words what dimensions give minimum outside surface area)?
No top?
Area=s^2+ h*s
dA/ds=2s+h=0
s=h/2
but volume=s^2*h
h=32/s^2
s=h/2=16/s^2
s=cubrt16
h=2cubrt16
check: volume=s^2*h=cubrt16 ^2 2cubrt16
volume=32
let the base be x by x ft and the height be h ft
Volume = x^2h
32 = x^2 h
h = 32/x^2
SA = x^2 + 4xh
= x^2 + 4x(32/x^2)
= x^2 + 128/x
d(SA)/dx = 2x - 128/x^2
= 0 for a max/min of SA
2x = 128/x^2
2x^3= 128
x^3 = 64
x = 4
then h = 32/16 = 2
so the box should have a base of 4ft by 4ft and a height of 2 ft
check:
V = 4x4x2 =32
SA = x^2 + 4xh = 16 + 32 = 48
let x = 3.9
h = 32/3.9^2
SA = 3.9^2 + 128/3.9 = 48.03 > 48
let x = 4.1
SA = 4.1^2 + 128/4.1 = 48.03 > 48
Here is the SA graph as shown by Wolfram,
notice that 48 is the minimum y value for x>0
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+128%2Fx
To find the dimensions that use the least amount of material, we need to minimize the outside surface area of the box.
Let's assume that the length of one side of the square base is "s" and the height of the box is "h".
The volume of the box is given as 32 ft^3, so we have the equation:
Volume = s^2 * h = 32
To minimize the surface area, we need to find the minimum value of the function that represents the surface area.
The surface area of the box can be calculated as follows:
Surface area = 2s^2 + 4sh
We can express "s" in terms of "h" using the volume equation:
s^2 = 32 / h
Substituting this expression for "s^2" into the surface area equation, we get:
Surface area = 2 * (32 / h) + 4sh
Surface area = (64 / h) + 4sh
To minimize the surface area, we can take the derivative of the surface area equation with respect to "h" and set it equal to zero:
d(Surface area) / d(h) = -64 / h^2 + 4s = 0
Solving this equation for "s", we get:
s = 16 / h^2
Substituting this expression for "s" back into the volume equation, we have:
(16 / h^2)^2 * h = 32
Simplifying this equation, we get:
256 / h^3 = 32
Solving for "h", we find:
h^3 = 8
Taking the cube root of both sides, we get:
h = 2 ft
Now that we have the value for "h", we can substitute it back into the expression for "s" to find its value:
s = 16 / (2^2)
s = 16 / 4
s = 4 ft
Therefore, the dimensions that use the least amount of material and result in a volume of 32 ft^3 are:
Length of one side of the square base = 4 ft
Height of the box = 2 ft
To find the dimensions that use the least amount of material (minimum outside surface area) for the given volume of the box, we can use mathematical optimization.
Let's denote the side length of the square base as 'x' and the height of the box as 'h'. Since the box has no top, the volume can be expressed as the product of the base area (x^2) and the height (h):
Volume = x^2 * h
Given that the volume is 32 ft³, we have the equation:
x^2 * h = 32
Now, to find the dimensions that minimize the outside surface area, we need to express the surface area in terms of 'x' and 'h'. The box has a square base, so the surface area of the base is x^2. Since the box has no top, there are four vertical sides; each side has an area equal to the base perimeter multiplied by the height. Therefore, the surface area of the four vertical sides is 4xh.
The total outside surface area is the sum of the surface area of the base and the four vertical sides:
Outside surface area = Base surface area + Vertical sides surface area
= x^2 + 4xh
To minimize the outside surface area while maintaining the volume of 32 ft³, we can use calculus optimization techniques. We need to minimize the outside surface area equation subject to the constraint equation (x^2 * h = 32).
First, we isolate 'h' in the constraint equation to express it in terms of 'x':
h = 32 / x^2
Next, substitute this expression for 'h' into the outside surface area equation:
Outside surface area = x^2 + 4x(32 / x^2)
= x^2 + 128 / x
Now, we differentiate the outside surface area equation with respect to 'x' to find the critical points:
d/dx (Outside surface area) = d/dx (x^2 + 128 / x)
= 2x - 128 / x^2
To find the critical points, set the derivative equal to zero and solve for 'x':
2x - 128 / x^2 = 0
2x^3 - 128 = 0
x^3 - 64 = 0
Now, factor the equation using the difference of cubes formula:
(x - 4)(x^2 + 4x + 16) = 0
From this equation, we have two possible solutions for 'x': x = 4, and the quadratic equation x^2 + 4x + 16 = 0, which has no real solutions.
Since we are looking for the dimensions of the box, we discard the quadratic equation solution since it does not make physical sense in this context.
Therefore, the only valid solution is x = 4. Now, substitute this value back into the constraint equation to find 'h':
h = 32 / x^2
= 32 / 4^2
= 32 / 16
= 2
So, the dimensions of the box that use the least amount of material (minimum outside surface area) for a volume of 32 ft³ are:
- Side length of the square base (x) = 4 ft
- Height of the box (h) = 2 ft