Question

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constructed by cutting four equal-sized surface corners from an 8-inch by 11.5 inch sheet of cardboard and folding up the sides.

1. Determine a function that relates the total surface area, s, (measured in square inches) of the open box to the size of the square cutout x (measured in inches).
2. What is the domain and range of the function s?
3. What is the surface area when a 1" x 1" square is cut out?
4. What size square cutout will result in a surface area of 20 in?
5. What is the surface area of the box when the volume is maximized? (Calculator)

Answer 1

A box designer has been charged with the task of determining the surface area of various open boxes (no lid) that can be constructed by cutting four equal-sized surface corners from an 8-inch by 11.5 inch sheet of cardboard and folding up the sides.

1. Determine a function that relates the total surface area, s, (measured in square inches) of the open box to the size of the square cutout x (measured in inches).

2. What is the domain and range of the function s?

3. What is the surface area when a 1" x 1" square is cut out?

4. What size square cutout will result in a surface area of 20 in?

5. What is the surface area of the box when the volume is maximized? (Calculator)

Explanation:

Answer 2

1)
`S = 2\cdot w\cdot l - 8\cdot x^(2)`, 2) The domain of S is
`0 \leq x \leq (√(w\cdot l))/(2)`. The range of S is
`0 \leq S \leq 2\cdot w \cdot l`, 3)
`S = 176\,in^(2)`, 4)
`x \approx 4.528\,in`, 5)
`S = 164.830\,in^(2)`

Explanation:

1) The function of the box is:

`S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)`

`S = 2\cdot w\cdot x - 4\cdot x^(2) + 2\cdot l\cdot x - 4\cdot x^(2) + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w`

`S = 2\cdot (w+l)\cdot x - 8\cdpt x^(2) + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x`

`S = 2\cdot w\cdot l - 8\cdot x^(2)`

2) The maximum cutout is:

`2\cdot w \cdot l - 8\cdot x^(2) = 0`

`w\cdot l - 4\cdot x^(2) = 0`

`4\cdot x^(2) = w\cdot l`

`x = (√(w\cdot l))/(2)`

The domain of S is
`0 \leq x \leq (√(w\cdot l))/(2)`. The range of S is
`0 \leq S \leq 2\cdot w \cdot l`

3) The surface area when a 1'' x 1'' square is cut out is:

`S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^(2)`

`S = 176\,in^(2)`

4) The size is found by solving the following second-order polynomial:

`20\,in^(2) = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^(2)`

`20\,in^(2) = 184\,in^(2) - 8\cdot x^(2)`

`8\cdot x^(2) - 164\,in^(2) = 0`

`x \approx 4.528\,in`

5) The equation of the box volume is:

`V = (w-2\cdot x)\cdot (l-2\cdot x) \cdot x`

`V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^(2)]\cdot x`

`V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^(2) + 4\cdot x^(3)`

`V = (8\,in)\cdot (11.5\,in)\cdot x - 2\cdot (19.5\,in)\cdot x^(2) + 4\cdot x^(3)`

`V = (92\,in^(2))\cdot x - (39\,in)\cdot x^(2) + 4\cdot x^(3)`

The first derivative of the function is:

`V' = 92\,in^(2) - (78\,in)\cdot x + 12\cdot x^(2)`

The critical points are determined by equalizing the derivative to zero:

`12\cdot x^(2)-(78\,in)\cdot x + 92\,in^(2) = 0`

`x_(1) \approx 4.952\,in`

`x_(2)\approx 1.548\,in`

The second derivative is found afterwards:

`V'' = 24\cdot x - 78\,in`

After evaluating each critical point, it follows that
`x_(1)` is an absolute minimum and
`x_(2)` is an absolute maximum. Hence, the value of the cutoff so that volume is maximized is:

`x \approx 1.548\,in`

The surface area of the box is:

`S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^(2)`

`S = 164.830\,in^(2)`