An open box is to be made from a 3ft by 8ft rectangular piece of sheet metal by cutting out squares of equal size from the four corners and bending up the sides. Find the maximum volume that the box c

To maximize the volume of the box, we want to choose the size of the squares to cut out such that the length and width of the base of the box are as large as possible. Let's call the length of the side of each square "x". When we cut out the squares, the dimensions of the rectangular piece of sheet metal left will be (8-2x) by (3-2x) (since we are removing x from both sides of the length and width). The height of the box will be x. So, the volume of the box will be V(x) = x(8-2x)(3-2x). To find the maximum volume, we can take the derivative of V(x), set it equal to zero, and solve for x: V'(x) = 12x^2 - 44x + 24 = 0 Solving for x using the \quadratic formula, we get x = 1 or x = 2. To determine which value of x gives the maximum volume, we can use the second derivative test: V''(x) = 24x - 44 V''(1) = -20, which means x = 1 is a local maximum. V''(2) = 4, which means x = 2 is a local minimum. Therefore, the maximum volume occurs when x = 1: V(1) = 1(8-2(1))(3-2(1)) = 12 ft^3 Rounding to the nearest integer, the maximum volume of the box is 12 ft^3.