Answer:
The company's maximum monthly profit is $18050
Explanation:
First we need to distribute everything in order to solve for the maximum
![f(x)=-2(x-10)(x-200)\\\\f(x)=-2(x^2-210x+2000)\\\\f(x)=-2x^2+420x-4000](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ki18w2rxhrfnkcw9lz37qy8xng5juq76f7.png)
Now we can use the equation
in order to find the x-value of the maximum
From our function, we know that a=-2 and b=420
We can plug in these values and solve for x to get
![x=(-420)/(-2(2)) \\\\x=(420)/(4) \\\\x=105](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hfvhl1gw85dfp3flaai554rx0m5oa28fdb.png)
Now we can plug our x-value into our function in order to find the maximum monthly profit
![f(x)=-2x^2+420x-4000\\\\f(105)=-2(105)^2+420(105)-4000\\\\f(105)=-2(11025)+44100-4000\\\\f(105)=-22050+44100-4000\\\\f(105)=18050](https://img.qammunity.org/2020/formulas/mathematics/middle-school/8w0pfuxck4ngja8m9sdsgqgmoo3y85io4h.png)