Answer:
The highest side to be cut out is 10/3 cm
Step-by-step explanation:
Let the size of the square be s, then the volume of the box is:
s(15 - 2s)(40 - 2s). This can be written as:

Taking the derivative, we have:

Set the above = 0, and take the roots:

s = 15 is not practical, so we use s = 10/3
The maximum volume is therefore;
![\begin{gathered} ((10)/(3))(15-(20)/(3))(40-(20)/(3)) \\ \\ =925.925\operatorname{cm}^3 \end{gathered}]()