HELP!!!!!!!!!!!!!

Question

You are making an open box from a rectangular sheet of cardboard by cutting squares 2

inches in length from each corner and folding up the sides. The length of the sheet of cardboard is 8 inches more than the width.

a. Write a polynomial expression that represents the

the total volume of the open box.

b. Find the volume of the open box when the width of the sheet of cardboard is 6 inches.

Answer 1

You're starting off with a rectangular cardboard with length = W + 8 inches and width = W inches. You're going to cut 2" squares out of each of the 4 corners and fold up the sides, so the sides of the resulting box will be length = W + 8 - 2 - 2 = W + 4. The width of the box is W - 2 - 2 = W - 4.

So the area of the base is l x w = (W+4)(W - 4) = W² + 4W- 4W - 16, or W² - 16.

a. Volume of box = area x height = 2(W² - 16) = 2W² - 32 in³ where W is the width of the original piece of cardboard.

b. Volume if W = 6". 2(6²) - 32 = 2(36) - 32 = 72 - 32 = 40 in³

Answer 2

L=8+W

Volume

= V =length x width x 2, after cutting 4 inches off the paper's Length and 4 inches off the paper's Width

= (L-4)(W-4)2

= (8+W-4)(W-4)2

= (W+4)(W-4)2

= 2W^2-32

V(W) =2W^2 -32

V(6) = 2(36) -32 = 72=32= 40 in^2

Volume 40 cubic inches when Width = 6 inches

or you could have found V as a function of L

V=(L-4)(W-4)2

= (L-4)(L-8-4)2

= (L-4)(L-12)2

= 2L^2 -32L +96

when W=6, L=6+8=14

V(L) = 2L^2-32L+96

V(14) = (2)14^2 -32(14)+ 96

= (2)196 - 448 + 96

= 392-448+96

= 40 cubic inches when Width = 6 inches