A box with a square base and open top must have a volume of 4,000 cm^3. Find the dimensions of the box that minimize the amount of material used.
sides of base cm
height cm
For sides of base I got 20cm and for height I got 10 cm? Anybody agree or disagree with me?
the volume
v = x^2 z = 4000, so z = 4000/x^2
the area is
a = x^2 + 4xz = x^2 + 16000/x
da/dx = 2x - 16000/x^2
da/dx=0 when
2x - 16000/x^2 = 0
x^3 = 8000
x = 20
So the box is 20x20x10 cm
Same as what I got
To minimize the amount of material used, we need to find the dimensions that minimize the surface area of the box.
Let's let the side length of the square base be "s" and the height of the box be "h."
The volume of the box is given as 4,000 cm^3, so we have the equation:
Volume = s^2 * h = 4,000 cm^3
We want to minimize the surface area of the box, so we need to find an equation for the surface area in terms of "s" and "h."
The surface area of the box consists of the area of the four sides of the base (4 * s^2) and the area of the open top (s^2). Since there is no material used for the bottom of the box, we don't include it in the surface area calculation.
So, the surface area equation is:
Surface Area = 4 * s^2 + s^2 = 5 * s^2
We can rewrite the volume equation in terms of "h":
h = 4,000 / (s^2)
Now, we can substitute the expression for "h" into the surface area equation:
Surface Area = 5 * s^2
To minimize the surface area, we need to find the minimum point of this quadratic equation.
Taking the derivative of the surface area equation with respect to "s," we get:
d(Surface Area)/ds = 10s
To find the minimum point, we set the derivative equal to zero:
10s = 0
s = 0
Since the side length cannot be zero, we discard that solution.
To find the dimensions that minimize the amount of material used, we need to solve for "s" and "h" using the volume equation.
Substituting the volume equation into the surface area equation, we have:
Surface Area = 5 * s^2 = 5 * (4,000 / h)
Since the two sides of the equation must be equal, we can solve for "h" in terms of "s":
s^2 = 4,000 / h
h = 4,000 / s^2
Now, we can substitute this expression for "h" back into the volume equation:
s^2 * (4,000 / s^2) = 4,000
This simplifies to:
4,000 = 4,000
Since the equation is satisfied for any value of "s," we can conclude that the box can have any dimensions as long as the volume is 4,000 cm^3. Therefore, there is no unique solution for the dimensions that minimize the amount of material used.