Maximize Volume of a Box (Maksimalkan Volume Balok)

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. 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.