Consider the following problem: A box with an open top is to be constructed from

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 comers 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 both x and y. (d) Use the given information to write an equation that relates the variables x and y. (e) Use part (d) to write the volume as a function of only x. V(x) = (f) Finish solving the problem by finding the largest volume that such a box can have. V =

Explanation / Answer

The cardboard is square

so the dimensions of the box are:

side length y = 3 - 2x ... probably better not to even use y b/c it's not required ... but they want it so there you go

and height = x

Volume = area of base x height

V = xy²

so V = x (3 - 2x)² ... [b/c y = 3 - 2x]

V = x(9 - 12x + 4x²)

V = 9x - 12x² + 4x³

V' = 9 - 24x + 12x² and V" = -24 + 24x

V' = 0 when 9 - 24x + 12x² = 0

4x² - 8x + 3 = 0

factoring:

(2x - 1)(2x - 3) = 0

2x - 1 = 0 or 2x - 3 = 0

x = 1/2 or x = 3/2

The cardboard is 3 ft square ... so 0 x 3/2 is the interval covering all the possible values of x (the side length of the cut out bits)

when x = 0, V = 0 - 0 + 0 = 0

when x = 3/2, V = 27/2 - 27 + 27/2 = 0

when x = 1/2, V = 2


so the absolute maximum is 2

and the largest possible volume is 2 ft³

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---