Answer :
To find the dimensions of the box that minimize the amount of material used, we need to minimize the surface area given the fixed volume. Since the box has a square base, let's denote the width of the square base by [tex]\( x \)[/tex] (in cm) and the height of the box by [tex]\( y \)[/tex] (in cm).
The volume [tex]\( V \)[/tex] of the box is given by the product of the area of the base and its height:
[tex]\[ V = x^2y \][/tex]
We know that the volume of the box needs to be 24,000 cm³, so we have the equation:
[tex]\[ x^2y = 24,000 \][/tex]
The surface area [tex]\( A \)[/tex] we want to minimize is the area of the base plus the area of the four sides of the box:
[tex]\[ A = x^2 + 4xy \][/tex]
Since [tex]\( y \)[/tex] can be expressed in terms of [tex]\( x \)[/tex] from the volume equation, we can substitute [tex]\( y \)[/tex] in the surface area equation:
[tex]\[ y = \frac{24,000}{x^2} \][/tex]
Plugging this into the surface area equation, we get:
[tex]\[ A(x) = x^2 + 4x\left(\frac{24,000}{x^2}\right) \][/tex]
[tex]\[ A(x) = x^2 + \frac{96,000}{x} \][/tex]
To minimize [tex]\( A(x) \)[/tex], we can take the derivative with respect to [tex]\( x \)[/tex] and set it equal to zero. This is because at a minimum point, the slope of the tangent to the curve [tex]\( A(x) \)[/tex] is zero.
Taking the derivative [tex]\( A'(x) \)[/tex] with respect to [tex]\( x \)[/tex], we get:
[tex]\[ A'(x) = 2x - \frac{96,000}{x^2} \times (-1) \][/tex]
[tex]\[ A'(x) = 2x + \frac{96,000}{x^2} \][/tex]
Setting the derivative equal to zero:
[tex]\[ 0 = 2x + \frac{96,000}{x^2} \][/tex]
Multiplying by [tex]\( x^2 \)[/tex] to clear the fraction:
[tex]\[ 0 = 2x^3 + 96,000 \][/tex]
Subtract 96,000 from both sides:
[tex]\[ -96,000 = 2x^3 \][/tex]
Divide by 2:
[tex]\[ -48,000 = x^3 \][/tex]
Find the cube root (note that only the positive root makes sense in this physical context, as dimensions cannot be negative):
[tex]\[ x = \sqrt[3]{48,000} \][/tex]
Approximate the cube root of 48,000 to get the dimension of the square base:
[tex]\[ x \approx 36.3 \text{ cm} \][/tex]
Now, we need to solve for [tex]\( y \)[/tex] using the earlier equation, and using the value we found for [tex]\( x \)[/tex]:
[tex]\[ y = \frac{24,000}{x^2} \][/tex]
[tex]\[ y = \frac{24,000}{36.3^2} \][/tex]
[tex]\[ y \approx \frac{24,000}{1,318.49} \][/tex]
[tex]\[ y \approx 18.2 \text{ cm} \][/tex]
Therefore, the dimensions of the box that minimize the amount of material used are approximately [tex]\( x = 36.3 \)[/tex] cm for the width of the square base and [tex]\( y = 18.2 \)[/tex] cm for the height of the box.
The volume [tex]\( V \)[/tex] of the box is given by the product of the area of the base and its height:
[tex]\[ V = x^2y \][/tex]
We know that the volume of the box needs to be 24,000 cm³, so we have the equation:
[tex]\[ x^2y = 24,000 \][/tex]
The surface area [tex]\( A \)[/tex] we want to minimize is the area of the base plus the area of the four sides of the box:
[tex]\[ A = x^2 + 4xy \][/tex]
Since [tex]\( y \)[/tex] can be expressed in terms of [tex]\( x \)[/tex] from the volume equation, we can substitute [tex]\( y \)[/tex] in the surface area equation:
[tex]\[ y = \frac{24,000}{x^2} \][/tex]
Plugging this into the surface area equation, we get:
[tex]\[ A(x) = x^2 + 4x\left(\frac{24,000}{x^2}\right) \][/tex]
[tex]\[ A(x) = x^2 + \frac{96,000}{x} \][/tex]
To minimize [tex]\( A(x) \)[/tex], we can take the derivative with respect to [tex]\( x \)[/tex] and set it equal to zero. This is because at a minimum point, the slope of the tangent to the curve [tex]\( A(x) \)[/tex] is zero.
Taking the derivative [tex]\( A'(x) \)[/tex] with respect to [tex]\( x \)[/tex], we get:
[tex]\[ A'(x) = 2x - \frac{96,000}{x^2} \times (-1) \][/tex]
[tex]\[ A'(x) = 2x + \frac{96,000}{x^2} \][/tex]
Setting the derivative equal to zero:
[tex]\[ 0 = 2x + \frac{96,000}{x^2} \][/tex]
Multiplying by [tex]\( x^2 \)[/tex] to clear the fraction:
[tex]\[ 0 = 2x^3 + 96,000 \][/tex]
Subtract 96,000 from both sides:
[tex]\[ -96,000 = 2x^3 \][/tex]
Divide by 2:
[tex]\[ -48,000 = x^3 \][/tex]
Find the cube root (note that only the positive root makes sense in this physical context, as dimensions cannot be negative):
[tex]\[ x = \sqrt[3]{48,000} \][/tex]
Approximate the cube root of 48,000 to get the dimension of the square base:
[tex]\[ x \approx 36.3 \text{ cm} \][/tex]
Now, we need to solve for [tex]\( y \)[/tex] using the earlier equation, and using the value we found for [tex]\( x \)[/tex]:
[tex]\[ y = \frac{24,000}{x^2} \][/tex]
[tex]\[ y = \frac{24,000}{36.3^2} \][/tex]
[tex]\[ y \approx \frac{24,000}{1,318.49} \][/tex]
[tex]\[ y \approx 18.2 \text{ cm} \][/tex]
Therefore, the dimensions of the box that minimize the amount of material used are approximately [tex]\( x = 36.3 \)[/tex] cm for the width of the square base and [tex]\( y = 18.2 \)[/tex] cm for the height of the box.
Explanation:
Variables:
s: side length of the square base (in cm)
h: height of the box (in cm)
Material Used:
Area of the base: s² (square centimeters)
Area of each side: sh (square centimeters) (There are 4 sides to consider)
Total surface area (excluding top): s² + 4sh
Volume Constraint:
The volume of the box is given as 24,000 cm³. We can express the volume in terms of our variables:
Volume = s²h = 24,000 cm³
Minimizing Material Usage:
We want to minimize the total surface area (excluding the top) while maintaining the required volume. Since we can't directly minimize the surface area, we can express it in terms of the volume using the volume constraint.
Here's how we can proceed:
Solve the volume equation for h: h = 24,000 cm³ / s²
Substitute this expression for h in the surface area formula: s² + 4s (24,000 cm³ / s²)
Simplify the combined expression: s² + 96,000 cm³ / s
Optimization:
This formula (s² + 96,000 cm³ / s) represents the surface area as a function of the base side length (s). To minimize the surface area, we need to find the minimum value of this function. This can be done through calculus (specifically taking the derivative and finding the minimum point).
However, there's a shortcut we can use:
We see that s² and 1/s terms are present in the formula. The product of these terms will always be 1 (s² * 1/s = 1). This suggests that the function might reach its minimum when these terms balance each other out.
Following the shortcut:
Set s² equal to 1/s: s⁴ = 1
Solve for s: s = ¹√∛1 (cube root of 1) = 1 (since any number raised to the power of 1 is itself)
Finding the Height (h):
Now that we know the base side length (s = 1 cm), we can find the height (h) using the volume constraint:
h = 24,000 cm³ / (1 cm)² = 24,000 cm³
Therefore, the dimensions of the box that minimize the material used are:
Base side length (s) = 1 cm
Height (h) = 24,000 cm³
