Amy's project in metal shop class is to make a box (with no lid). Her shop teacher gave her a piece of metal that is 28 cm long and 20 cm wide. The assignment is to form the box by cutting squares out of each corner and bending the sides as shown below. The shop teacher, Mr. Rust, will give extra credit to any student who forms the box that holds the most when the cut-out square is a whole number of centimeters on each side. Amy wants the extra point! How large a square should she cut from each corner to form the box with the greatest volume?
Let x = side of square to cut off from each corner
20 - 2x = width of bottom rectangle
28 - 2x = length of bottom rectangle
V = x(20 - 2x)(28 - 2x) = 4x(10 - x)(14 -x) =
V = 4x(140 - 24x + x²)
V = 4(140x - 24x² + x³)
Differentiating and equating to zero
dV/dx = 4(140 - 48x + 3x²) = 0
3x² - 48x + 140 = 0
x = {48 ± √[48² - 4(3)(140)] } /2(3)
x = {48 ± √624] } /6
x = {48 ± 4√39] } /6
x = (24 ± 2√39) / 3
x = (24 + 2√39) / 3 = 12.16 Discard, > ½20
x = (24 - 2√39) / 3 = 3.8 inch
Use x = 4 since whole number of cm is required.
So each square to cut is 4 cm on the side.
4 cm squares gives a box of 4 x 12 x 20 = 960 cm^3
To find out how large a square Amy should cut from each corner to form the box with the greatest volume, we need to follow a set of steps:
1. Let's assume each side of the square to be cut out has a length of "x" cm.
2. When the square is cut out from each corner, the resulting dimensions of the metal sheet will be (28 - 2x) cm and (20 - 2x) cm.
3. The volume of the box can be calculated by multiplying the length, width, and height. The height, in this case, will be the length of the square cut out from each corner (x).
4. Therefore, the volume of the box is V = x(28 - 2x)(20 - 2x).
Now, to find the value of "x" that maximizes the volume of the box, we can use a mathematical technique called differentiation. We differentiate the volume function with respect to "x" and find the critical points.
Let's differentiate the volume function:
dV/dx = (28 - 4x)(20 - 2x) + x(-4)(20 - 2x) + x(28 - 4x)(-2)
Simplifying further:
dV/dx = (28 - 4x)(20 - 2x) - 4x(20 - 2x) - 2x(28 - 4x)
Expanding terms:
dV/dx = 560 - 80x - 56x + 8x^2 - 80x + 8x^2 - 56x + 8x^3
Combining like terms:
dV/dx = 16x^3 - 192x + 560
Now, we set this derivative equal to zero to find critical points:
16x^3 - 192x + 560 = 0
To solve this cubic equation, we can use a numerical method or a graphing calculator to find the value of "x." After calculating "x," we substitute it back into the volume function to find the maximum volume of the box.
Alternatively, we could use a graphing calculator or an optimization software to directly find the value of "x" that maximizes the volume function.
Once we find the value of "x," Amy should cut squares with sides measuring that value from each corner of the metal sheet to form the box with the greatest volume and secure that extra credit from Mr. Rust.