Maximize Volume of a Box
Optimization Problem

How to maximize the volume of a box using the first derivative of the volume. Volume optimization problem with solution.

Problem 1: A sheet of metal 12 inches by 10 inches is to be used to make a open box. Squares of equal sides x are cut out of each corner then the sides are folded to make the box. Find the value of x that makes the volume maximum.

maximize volume problem 1

Solution to Problem 1:

  • We first use the formula of the volume of a rectangular box.
    V = L * W * H
  • The box to be made has the following dimensions:
    L = 12 - x
    W = 10 - 2x
    H = x
  • We now write the volume of the box to ba made as follows:
    V(x) = x (12 - 2x) (10 - 2x) = 4x (6 - x) (5 - x)
    = 4x (x 2 -11 x + 30)
  • We now determine the domain of function V(x). All dimemsions of the box must be positive or zero, hence the conditions
    x > = 0 and 6 - x > = 0 and 5 - x > = 0
  • Solve the above inequalities and find the intersection, hence the domain of function V(x)
    0 < = x < = 5
  • Let us now find the first derivative of V(x) using its last expression.
    dV / dx = 4 [ (x 2 -11 x + 3) + x (2x - 11) ]
    = 3 x 2 -22 x + 30
  • Let us now find all values of x that makes dV / dx = 0 by solving the quadratic equation
    3 x 2 -22 x + 30 = 0
  • Two values make dV / dx = 0: x = 5.52 and x = 1.81, rounded to one decimal place. x = 5.52 is outside the domain and is therefore rejected. Let us now examine the values of V(x) at x = 1.81 and the endpoints of the domain.
    V(0) = 0 , v(5) = 0 and V(1.81) = 96.77 (rounded to two decimal places)
  • So V(x) is maximum for x = 1.81 inches. The graph of function V(x) is shown below and we can clearly see that there is a maximum very close to 1.8.
    graph of V(x), problem 1

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