The given polynomial is
![p(x)=-0.002x^2+2.5x-600](https://img.qammunity.org/2023/formulas/mathematics/college/53kozfw36jjkvmpnyiexhn8l5cs64pphor.png)
Differentiate with respect to x, we get
![p^(\prime)(x)=2(-0.002)x+2.5](https://img.qammunity.org/2023/formulas/mathematics/college/c0rovixs908ca76s3c42mlp7n18j6ou54s.png)
![p^(\prime)(x)=-0.002x+2.5](https://img.qammunity.org/2023/formulas/mathematics/college/kd19tqk2ybyoyjjb39ppr0nr3crr8v844m.png)
Equate this to zero, we get
![-0.004x+2.5=0](https://img.qammunity.org/2023/formulas/mathematics/college/y92z5jsnglxldovq60720tzb49vgzg8q1n.png)
![-0.004x=-2.5](https://img.qammunity.org/2023/formulas/mathematics/college/nbue38k48d5ocbarv1fizktdguko3g1uyg.png)
![0.004x=2.5](https://img.qammunity.org/2023/formulas/mathematics/college/151rd1fu43i43ytqe4vxq44i2y6sjd4cga.png)
multiply by 1000 on both sides, we get
![0.004x*1000=2.5*1000](https://img.qammunity.org/2023/formulas/mathematics/college/isxwonl2oi5pywuu86t4fmdy45q20v8bbq.png)
![4x=2500](https://img.qammunity.org/2023/formulas/mathematics/college/k1nwr9fcp608txcs6qmverw6erp4tmusob.png)
![x=(2500)/(4)=625](https://img.qammunity.org/2023/formulas/mathematics/college/icmqmonmpithc7y2chvzru3ro47mczuhya.png)
Hence they must sell 625 patterns to attain maximum profit.
Substitute x=625 in the given equation p(x), we get
![P(625)=-0.002(625)^2+2.5(625)-600](https://img.qammunity.org/2023/formulas/mathematics/college/44czknbpnm5jnstden9lseg4klt7qu7n4b.png)
![=-0.002*390625+2.5*625-600](https://img.qammunity.org/2023/formulas/mathematics/college/z1ky4pjeavqgryi813b0abou237ivu0u9t.png)
![=-781.25+1562.5-600](https://img.qammunity.org/2023/formulas/mathematics/college/hg4m6t84u7c8nl6914h5lki788by8mwusb.png)
![P(625)=181.25](https://img.qammunity.org/2023/formulas/mathematics/college/ckht4utwcq38ygyxcwqzjuvg0lfy9tiuxq.png)
Hence the maximum profit is $ 181.25.