Question

PLZ SHOW ALL WORK

Sabrina is making an open box from a piece of cardboard that has a width of 12 inches and a length of 18 inches. She’ll form the box by making cuts at the corners and folding up the sides so that the flaps are square. If she wants the box to have a volume of 224 in3, how long should she make the cuts?


Part I: Given what you know about the box Sabrina is making, write a polynomial to represent each dimension (width, length, and height) of the finished box. (3 points)

Width =

Length =

Height =

Part II: The formula for volume is v = l • w • h. Write an equation to represent the volume of the box. Simplify on the right side by multiplying the polynomials and writing the answer in descending order. Show your work.

HINT: First multiply the binomials representing length and width. Next, multiply the resulting trinomial by the expression representing height. (3 points)

v =










Part III: The length of each cut at the corners is represented by x. To make a box, x must fall in a certain range of values. For example, x cannot be 15 because that would make the flap longer than the width of the box. Identify one reasonable value of x. Why did you choose that value? (1 point)

Answer 1

Part I
We have the size of the sheet of cardboard and we'll use the variable "x" to represent the length of the cuts. For any given cut, the available distance is reduced by twice the length of the cut. So we can create the following equations for length, width, and height.
width: w = 12 - 2x
length: l = 18 - 2x
height: h = x

Part II
v = l * w * h
v = (18 - 2x)(12 - 2x)x
v = (216 - 36x - 24x + 4x^2)x
v = (216 - 60x + 4x^2)x
v = 216x - 60x^2 + 4x^3
v = 4x^3 - 60x^2 + 216x

Part III
The length of the cut has to be greater than 0 and less than half the length of the smallest dimension of the cardboard (after all, there has to be something left over after cutting out the corners). So 0 < x < 6

Let's try to figure out an x that gives a volume of 224 in^3. Since this is high school math, it's unlikely that you've been taught how to handle cubic equations, so let's instead look at integer values of x. If we use a value of 1, we get a volume of:
v = 4x^3 - 60x^2 + 216x
v = 4*1^3 - 60*1^2 + 216*1
v = 4*1 - 60*1 + 216
v = 4 - 60 + 216
v = 160

Too small, so let's try 2.
v = 4x^3 - 60x^2 + 216x
v = 4*2^3 - 60*2^2 + 216*2
v = 4*8 - 60*4 + 216*2
v = 32 - 240 + 432
v = 224

And that's the desired volume.
So let's choose a value of x=2.
Reason?
It meets the inequality of 0 < x < 6 and it also gives the desired volume of 224 cubic inches.