Question

Without cover a box is formed by 6 m x 16 m rectangular steel sheet on cutting the squares of length x m from its each corner. Then find the maximum volume of the box.

Hint:

Always establish the physical constraints (domain) of the variable in optimization problems. This often helps in discarding extraneous mathematical solutions obtained from setting the derivative to zero.

Answer 1

Step 1: Understanding the Concept:
This is an optimization problem where we need to find the maximum volume of an open-top box. The volume is expressed as a function of the side 'x' of the square cutouts, and calculus is used to find the maximum value of this function.
Step 2: Key Formula or Approach:
1. Express the dimensions (length, width, height) of the box in terms of x.
2. Formulate 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 = 16 m and W = 6 m.
When squares of side x are cut from each corner, the dimensions of the folded box become:
Height: \( h = x \)
Length: \( l = 16 - 2x \)
Width: \( w = 6 - 2x \)
The volume \( V(x) \) is given by \( l \cdot w \cdot h \): \[ V(x) = (16 - 2x)(6 - 2x)x = (96 - 32x - 12x + 4x^2)x \] \[ V(x) = 4x^3 - 44x^2 + 96x \] For the dimensions to be physically possible, \( x>0 \), \( 6 - 2x>0 \Rightarrow x<3 \), and \( 16 - 2x>0 \Rightarrow x<8 \). The valid domain for x is \( 0<x<3 \).
Find the first derivative to locate critical points: \[ V'(x) = 12x^2 - 88x + 96 \] Set \( V'(x) = 0 \): \[ 12x^2 - 88x + 96 = 0 \] Divide by 4: \[ 3x^2 - 22x + 24 = 0 \] Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{22 \pm \sqrt{(-22)^2 - 4(3)(24)}}{2(3)} = \frac{22 \pm \sqrt{484 - 288}}{6} = \frac{22 \pm \sqrt{196}}{6} = \frac{22 \pm 14}{6} \] This gives two possible values: \( x = \frac{36}{6} = 6 \) and \( x = \frac{8}{6} = \frac{4}{3} \).
Within our domain \( 0<x<3 \), the only valid critical point is \( x = \frac{4}{3} \).
Use the second derivative test to confirm it's a maximum: \[ V''(x) = 24x - 88 \] \[ V''\left(\frac{4}{3}\right) = 24\left(\frac{4}{3}\right) - 88 = 32 - 88 = -56 \] Since \( V''<0 \), the volume is maximized at \( x = \frac{4}{3} \).
Calculate the maximum volume: \[ V\left(\frac{4}{3}\right) = \left(16 - 2\left(\frac{4}{3}\right)\right)\left(6 - 2\left(\frac{4}{3}\right)\right)\left(\frac{4}{3}\right) = \left(\frac{48-8}{3}\right)\left(\frac{18-8}{3}\right)\left(\frac{4}{3}\right) \] \[ V\left(\frac{4}{3}\right) = \left(\frac{40}{3}\right)\left(\frac{10}{3}\right)\left(\frac{4}{3}\right) = \frac{1600}{27} \] Step 4: Final Answer:
The maximum volume of the box is \( \frac{1600}{27} \) cubic meters.