Question

Consider the following problem:  A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides.  Find the largest volume that such a box can have.

    (a)  Draw several diagrams to illustrate the situation, some short boxes with large bases and some tall boxes with small bases.  Find the volumes of several such boxes.

    (b)  Draw a diagram illustrating the general situation.  Let x denote the length of the side of the square being cut out.  Let y denote the length of the base.

    (c)  Write an expression for the volume V in terms of x and y.
V =     

1

    (d)  Use the given information to write an equation that relates the variables x and y.

2

    (e)  Use part (d) to write the volume as a function of x.
V(x) =     

3

(f)  Finish solving the problem by finding the largest volume that such a box can have.
V = 4 ft³

Answer 1

1 ft 0.25 ft 0.25 lft 15 ft (b)Let the side of the corner be a and the side of the base be b

The side of the corner is a ft Then height of the box - a ft And the side of the square base is b Volume of the box, V - lengthx widthx height (d)Since total length of the board Then b 3-2a ft 3 ft (e) Since b- 3-2a then V3-2a(3 2axa ft -a(9-12a+4d2 V-9a-12a +4a....1)

The volume of the box is V-9a-12a2 +4a3 Differentiate with respect to a V 9-24a+12a2 Now V'=0 12a2 -24a+9 0 4a2 -8a +3-0 4a2-2a-6a + 3:0 (2a -3)(2a-1)-0 a__ or a=- 2 Second differentiation of V V"24 + 24a V"(3/2)--24 +36 12 0 And V"-(1/2)--24 + 24(1/2)--12-0

So by the second derivative test, volume will be maximum when a -ft 2 And the maximum volume-k3-1-2n]from (1) 2