Question

A box is formed by a 3 m x 8 m rectangular steel-sheet on cutting the squares of length x m from its each corner to form the box without cover. Then find the maximum volume of the box so formed.

Hint:

In optimization problems involving geometric shapes, always determine the valid domain for the variable first. This can quickly eliminate extraneous solutions found from the derivative, saving time and preventing errors.

Answer 1

Step 1: Understanding the Concept:
This is an optimization problem that requires the use of differential calculus. We need to express the volume of the box as a function of the side length 'x' of the cut-out squares, and then find the value of 'x' that maximizes this volume.
Step 2: Key Formula or Approach:
1. Determine the dimensions (length, width, height) of the resulting open box in terms of x.
2. Write the volume function \( V(x) \).
3. Find the derivative \( V'(x) \) and set it to zero to find critical points.
4. Use the second derivative test, \( V''(x) \), to confirm that the critical point corresponds to a maximum.
Step 3: Detailed Explanation:
The original sheet has dimensions L = 8 m and W = 3 m.
When squares of side x are cut from each corner, the sheet is folded up. The dimensions of the resulting box will be:
- Height: \( h = x \) - Length: \( l = 8 - 2x \) - Width: \( w = 3 - 2x \) The volume of the box is \( V(x) = l \cdot w \cdot h \). \[ V(x) = (8 - 2x)(3 - 2x)x \] \[ V(x) = (24 - 16x - 6x + 4x^2)x \] \[ V(x) = 4x^3 - 22x^2 + 24x \] For the dimensions to be positive, we must have \( x>0 \), \( 3 - 2x>0 \implies x<1.5 \), and \( 8 - 2x>0 \implies x<4 \). The valid domain for x is \( 0<x<1.5 \).
Now, we find the first derivative to find critical points: \[ V'(x) = \frac{dV}{dx} = 12x^2 - 44x + 24 \] Set \( V'(x) = 0 \): \[ 12x^2 - 44x + 24 = 0 \] Divide by 4: \[ 3x^2 - 11x + 6 = 0 \] Factor the quadratic equation: \[ (3x - 2)(x - 3) = 0 \] The possible values for x are \( x = \frac{2}{3} \) or \( x = 3 \).
From our domain \( 0<x<1.5 \), the only valid solution is \( x = \frac{2}{3} \).
To confirm this is a maximum, we use the second derivative test: \[ V''(x) = 24x - 44 \] \[ V''\left(\frac{2}{3}\right) = 24\left(\frac{2}{3}\right) - 44 = 16 - 44 = -28 \] Since \( V''\left(\frac{2}{3}\right)<0 \), the volume is maximum at \( x = \frac{2}{3} \).
Calculate the maximum volume: \[ V\left(\frac{2}{3}\right) = 4\left(\frac{2}{3}\right)^3 - 22\left(\frac{2}{3}\right)^2 + 24\left(\frac{2}{3}\right) \] \[ = 4\left(\frac{8}{27}\right) - 22\left(\frac{4}{9}\right) + 16 \] \[ = \frac{32}{27} - \frac{88}{9} + 16 = \frac{32 - 264 + 432}{27} = \frac{200}{27} \] Step 4: Final Answer:
The maximum volume of the box is \( \frac{200}{27} \) cubic meters.