Cuboid
Anya's box will have larger volume as compared to Terence's box
x cannot be 0 as length of square cannot be 0.
x cannot be 4.25 and has to be less than 4.25 so that there should be enough breadth of cardboard to make the box (Height of box cannot be 0)
Step-by-step explanation:
Length l = 11 in and breadth b= 8.5 in
Square of equal sizes cut from 4 corners.Let the side of square is x
so l = x + x + y where y is remaining length, b = x + x + z where z is remaining breadth.
The resultant box after cutting the corner squares will have length as y and breadth as z and height will x
In case of Anya, x = 1.5 in
![l = 1.5 + 1.5 + y\\11 = 3 + y\\y = 11 - 3 = 8\\](https://tex.z-dn.net/?f=l%20%3D%201.5%20%2B%201.5%20%2B%20y%5C%5C11%20%3D%203%20%2B%20y%5C%5Cy%20%3D%2011%20-%203%20%3D%208%5C%5C)
![b = 1.5 + 1.5 + z\\8.5 = 3 + z\\z = 8.5 - 3\\z = 5.5\\](https://tex.z-dn.net/?f=b%20%3D%201.5%20%2B%201.5%20%2B%20z%5C%5C8.5%20%3D%203%20%2B%20z%5C%5Cz%20%3D%208.5%20-%203%5C%5Cz%20%3D%205.5%5C%5C)
so volume for Anya = V
⇒
= 1.5 × 8 × 5.5 = 66 cubic inches
In case of Terence, x = 3 in
![l = 3 + 3 + y\\11 = 6 + y\\y = 11 - 6 = 5\\\\](https://tex.z-dn.net/?f=l%20%3D%203%20%2B%203%20%2B%20y%5C%5C11%20%3D%206%20%2B%20y%5C%5Cy%20%3D%2011%20-%206%20%3D%205%5C%5C%5C%5C)
![b = 3 + 3 + z\\8.5 = 6 + z\\z = 8.5 - 6\\z = 2.5\\](https://tex.z-dn.net/?f=b%20%3D%203%20%2B%203%20%2B%20z%5C%5C8.5%20%3D%206%20%2B%20z%5C%5Cz%20%3D%208.5%20-%206%5C%5Cz%20%3D%202.5%5C%5C)
so volume for Terence= ![V_{t}](https://tex.z-dn.net/?f=V_%7Bt%7D)
⇒
= 3 × 5 × 2.5 = 37.5 cubic inches
Hence, Anya's box will have larger volume as compared to Terence's box
x cannot be 0 as length of square cannot be 0.
x cannot be 4.25 and has to be less than 4.25 so that there should be enough breadth of cardboard to make the box (Height of box cannot be 0)
So X is limited to lie between 0 and 4.25 in