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

The expression of volume in terms of x and y is [tex]V = x y^2[/tex].
The relationship between x and y is [tex]y = 3 - 2x[/tex],
Volume as a function of x is: [tex]V(x) =x \times (3 - 2x)^2[/tex].
The dimension that maximize the volume is: [tex]x = \frac 12[/tex] and [tex]y = 2[/tex]

(c) Expression for volume

The volume (V) of the box is:

[tex]V = x \times y \times y[/tex]

[tex]V = x y^2[/tex]

(d) Relationship between x and y

The length of the box is given as:

[tex]Length = 3[/tex]

The side of the box is constructed from the original square piece.

This means that:

[tex]y = Length - 2 \times x[/tex]

So, we have:

[tex]y = 3 - 2 \times x[/tex]

[tex]y = 3 - 2x[/tex]

(e) Volume as a function of x

In (c), we have:

[tex]V = x y^2[/tex]

Substitute [tex]y = 3 - 2x[/tex]

[tex]V =x \times (3 - 2x)^2[/tex]

So, we have:

[tex]V(x) =x \times (3 - 2x)^2[/tex]

(f) The largest dimension of the box

In (e), we have:

[tex]V(x) =x \times (3 - 2x)^2[/tex]

Expand

[tex]V(x) =x \times (9 - 12x + 4x^2)[/tex]

[tex]V(x) =9x - 12x^2 + 4x^3[/tex]

Differentiate

[tex]V'(x) =9 - 24x + 12x^2[/tex]

Equate to 0

[tex]9 - 24x + 12x^2 = 0[/tex]

Divide through by 3

[tex]3 - 8x + 4x^2 = 0[/tex]

Rewrite as:

[tex]4x^2 - 8x +3 = 0[/tex]

Expand

[tex]4x^2 - 6x - 2x +3 = 0[/tex]

Factorize

[tex]2x(2x - 3) -1( 2x -3 ) = 0[/tex]

Factor out 2x - 3

[tex](2x - 1) ( 2x -3 ) = 0[/tex]

Split

[tex]2x - 1 = 0\ or\ 2x -3 = 0[/tex]

Solve for x

[tex]2x = 1 \ or\ 2x =3[/tex]

Divide by 2

[tex]x = \frac12 \ or\ x =\frac 32[/tex]

Recall that:

[tex]y = 3 - 2x[/tex]

When [tex]x = \frac 12[/tex]

[tex]y = 3 - 2 \times \frac 12[/tex]

[tex]y = 2[/tex]

When [tex]x = \frac 32[/tex]

[tex]y = 3 - 2 \times \frac 32[/tex]

[tex]y = 0[/tex]

[tex]y = 0[/tex] is not a feasible solution

So, the dimensions that maximize the volume are:

[tex]x = \frac 12[/tex] and [tex]y = 2[/tex]