You are given a rectangular sheet of cardboard that measures 11 in. by 8.5 in. (see the diagram below). A small square of the sa

Question

22 days ago

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You are given a rectangular sheet of cardboard that measures 11 in. by 8.5 in. (see the diagram below). A small square of the sa

me size is cut from each corner, and each side folded up along the cuts to from a box with no lid.
1. Anya thinks the cut should be 1.5 inches to create the greatest volume, while Terrence thinks it should be 3 inches.

Explain how both students can determine the formula for the volume of the box.

Determine which student's suggestion would create the larger volume.

Explain how there can be two different volumes when each student starts with the same size cardboard.

2. Why is the value of x limited to 0 in. < x < 4.25 in.?

Mathematics

1 answer:

AnnZ [9K] · Answer 1 · 2026-03-05T20:47:51+03:00

Answer:

Terrence's

Step-by-step explanation:

The cut length of the square corresponds to the box height.

1. a

In Anya's approach: 8.5 - 1.5 = 7, 11 - 1.5 = 9.5; with the height being 1.5, the volume calculates as height x length x width = 1.5 x 9.5 x 7 = 99.75 square inches.

Terrence's method: 8.5 - 3 = 5.5, 11 - 3 = 8. Volume = 5.5 x 8 x 3 = 132 square inches. Clearly, 99.75 < 132 square inches, indicating Terrence's approach would yield a larger volume.

1. b

The dimensions of the box rely on the length, width, and height of the cardboard cutout, thus varying cutting measurements for the same cardboard size can produce different box dimensions.

2. Squares are cut from each corner, thus the combined size of the two squares on the cardboard must be less than the shorter cardboard side. The shorter side is 8.5 inches, which divided by 2 results in 4.25 inches, therefore squares cannot exceed this size. If squares are taken out measuring 4.25 inches, it will leave a strip 2.5 inches wide that cannot form a box.

To cut out squares of 5 inches, how you position them can cause overlap or extend beyond the cardboard. While 5 + 5 equals ten, making the longer side of 11 inches adequate, the 8.5 inches side allows zero room for the 10 inches.