From the problem, the length and the width will be reduced by twice the side of the square.
The length of the box will be :
![26-2x](https://img.qammunity.org/2023/formulas/mathematics/college/c4xazdbzreudnpw5xmwrn50mtoocwhe5i3.png)
The width of the box will be :
![20-2x](https://img.qammunity.org/2023/formulas/mathematics/college/q53tb93q746v2o0znc88lnvfjxk1hxbxn3.png)
and the height will be the measurement of the square side :
![x](https://img.qammunity.org/2023/formulas/mathematics/high-school/7i9rhkmy8weow049o4r221u9e7b2s5rdwo.png)
Note that the volume of a box is length x width x height.
1. The volume will be :
![V(x)=x(26-2x)(20-2x)](https://img.qammunity.org/2023/formulas/mathematics/college/q2m5ipr275ifkfgvcazxsr5xrngwoqkm1o.png)
Expand and simplify the function :
![\begin{gathered} V(x)=x(26-2x)(20-2x) \\ V(x)=x(520-40x-52x+4x^2) \\ V(x)=x(4x^2-92x+520) \\ V(x)=4x^3-92x^2+520x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/uy47i7n2hh0g8nmpvlf58cvw6lkwrvc0m3.png)
2. Graph the function using desmos.