the volume of an open topped box with a square base is 245 cubic centimeters. find the height in centimeters, of the box that uses the least amount of material
Answer is 3.942
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Well, the box with the least amount of material would be a paper-thin box, completely flat. That way, it would use the least amount of material possible! However, I'm guessing you're looking for a more practical answer. Let's solve it!
Let's call the length of each side of the square base "x", and the height of the box "h". The volume of the box is given as 245 cubic centimeters, so we have:
V = x^2 * h = 245
To minimize the amount of material, we need to minimize the surface area of the box. The surface area consists of the area of the square base plus the area of the four sides:
S = x^2 + 4xh
Now, we can express "x" in terms of "h" using the volume equation:
x^2 = 245/h
Substituting this into the surface area equation, we get:
S = (245/h) + 4h(245/h)
Simplifying further, we have:
S = 245/h + 980/h
S = (245 + 980)/h
S = 1225/h
To minimize the surface area, we need to maximize "h" or make it as large as possible. Therefore, the height that uses the least amount of material would be infinity! However, in practical terms, we usually work with finite measurements, so we need to be specific about the dimensions you're working with.
To find the height of the box that uses the least amount of material, we need to minimize the surface area of the box.
Let's denote the side length of the square base as "x" and the height of the box as "h".
The volume of the box is given as 245 cubic centimeters, so we have the equation:
V = x^2 * h = 245
To minimize the surface area, we need to find the derivative of the surface area function with respect to "h" and set it equal to zero.
The surface area of the box consists of the area of the square base and the four equal sides:
A = x^2 + 4xh
To find the derivative, we can differentiate the surface area function with respect to "h":
dA/dh = 4x
Setting the derivative equal to zero:
4x = 0
This implies that x = 0, which is not a valid solution. Therefore, the derivative does not provide any information about the minimum surface area.
Next, we can solve the equation for the volume in terms of "h":
h = 245 / x^2
Substituting this into the surface area equation:
A = x^2 + 4x (245 / x^2)
Simplifying:
A = x^2 + (980 / x)
To minimize the surface area, we can find the critical points by taking the derivative of the surface area function with respect to "x" and setting it equal to zero.
dA/dx = 2x - (980 / x^2)
Setting the derivative equal to zero:
2x - (980 / x^2) = 0
Multiplying through by x^2:
2x^3 - 980 = 0
Dividing through by 2:
x^3 - 490 = 0
To solve this equation, we can use the cube root:
x = ∛490 ≈ 7.85
Now, we can substitute this value of "x" back into the equation for the height:
h = 245 / x^2
h ≈ 245 / (7.85)^2
h ≈ 245 / 61.60
h ≈ 3.98
Therefore, the height of the box that uses the least amount of material is approximately 3.98 centimeters.
To find the height of the box that uses the least amount of material, we need to understand the relationship between the volume and the amount of material used.
First, let's define the variables:
- V: Volume of the box
- h: Height of the box
- s: Length of one side of the square base
The volume of an open-topped box with a square base is given by the formula:
V = s² * h
Given that the volume is 245 cubic centimeters, we can write the equation as:
245 = s² * h
To find the height that minimizes the amount of material used, we need to minimize the surface area of the box. Since the base is fixed as a square, we need to minimize the area of the remaining four sides.
The area of one side of the box is given by the formula:
A = s * h
We want to find the minimum surface area, which is equivalent to finding the minimum value of A.
To minimize A, we can use the calculus concept of finding the minimum of a function. Since we have two variables, we need to use the method of constrained optimization, where we express one variable in terms of the other and substitute it into the equation.
From the volume equation, we can express h in terms of s as:
h = 245 / s²
Substituting this expression for h into the area equation, we get:
A = s * (245 / s²)
Simplifying further:
A = 245 / s
To minimize A, we find the derivative of A with respect to s and set it equal to zero:
dA/ds = 0
Differentiating A with respect to s gives us:
dA/ds = -245 / s²
Setting dA/ds equal to zero, we have:
-245 / s² = 0
Solving for s, we find that s = 0 is the only solution. However, since s represents the length of a side, it cannot be zero. Therefore, there is no minimum value for A.
This means that the least amount of material is used when the height of the box is infinitely large. In practical terms, this means that the box can have any height greater than zero, and the material used will be minimized.
In summary, the height of the box that uses the least amount of material is achieved by making it infinitely tall.