Answer:
To find the lengths of the edges that give the minimum surface area for a closed rectangular box with a volume of 42 cm^3, we need to optimize the surface area of the box.
Let's call the three dimensions of the box x, y, and z. Since it's a closed rectangular box, all sides are perpendicular to each other.
The volume of the box is given by the formula V = x * y * z, which in this case is 42 cm^3.
To optimize the surface area, we need to minimize it. The surface area of the box is given by the formula A = 2xy + 2xz + 2yz.
To minimize the surface area, we can use calculus. Taking the derivative of the surface area formula with respect to one variable and setting it equal to zero will help us find the critical points.
Differentiating the surface area formula with respect to x, y, and z, we get:
dA/dx = 2y + 2z = 0,
dA/dy = 2x + 2z = 0,
dA/dz = 2x + 2y = 0.
Solving these equations, we find that x = y = -z.
Since the lengths cannot be negative, we can disregard the negative solution. Therefore, x = y = -z is not a valid solution.
To find the lengths of the edges giving the minimum surface area, we can try different values for x, y, and z, while keeping their product equal to 42 cm^3.
For example, if we let x = 7 cm, y = 3 cm, and z = 2 cm, their product is 7 * 3 * 2 = 42 cm^3, which satisfies the volume requirement.
Calculating the surface area using these values, we get A = 2(7)(3) + 2(7)(2) + 2(3)(2) = 42 + 28 + 12 = 82 cm^2.
We can try other combinations as well, such as x = 6 cm, y = 7 cm, and z = 1 cm, or x = 14 cm, y = 1 cm, and z = 3 cm, as long as their product is 42 cm^3.
In conclusion, there are multiple combinations of edge lengths that give the minimum surface area for a closed rectangular box with a volume of 42 cm^3. Examples include x = 7 cm, y = 3 cm, and z = 2 cm, or x = 6 cm, y = 7 cm, and z = 1 cm.