SOLUTION
From the what is given
![\begin{gathered} x+y\le5 \\ x\ge3 \\ 2y\le8 \\ x\ge0 \\ y\ge0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/77j54l4hgw6nme063wtvdlcqhpxx8jwsa2.png)
We have the graph as shown below
We are told that the MAX is
![5x+2y](https://img.qammunity.org/2023/formulas/mathematics/high-school/k24helfk641w2jl38vxl3g2aiqa4bdh7iy.png)
Substituting these required points into the equation, our maximum becomes
![\begin{gathered} 5x+2y \\ \text{For (3, 2)} \\ 5(3)+2(2)=15+4=19 \\ \text{For }(3,\text{ 0)} \\ 5(3)+2(0)=15+0=15 \\ \text{For (5, 0)} \\ 5(5)+2(0)=25+0=25 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/n11usv9myg4qb37vhigv90w4xx3vl3ld3k.png)
We can see that the maximum is 25 at for units of 5, that is x = 5
But we are told in (B) that
![x\ge3](https://img.qammunity.org/2023/formulas/mathematics/college/jntip3o839iy3s22fp98pssinijzwbjoan.png)
Hence the surplus unit is
![5-3=2](https://img.qammunity.org/2023/formulas/mathematics/high-school/kenkpos2v6szngwq9af5wiu5xvac1k1hi9.png)
Hence the answer is 2