The maximum units is 200 and , Total revenue is $8,000
Step-by-step explanation:
Here we have , A manufacturer finds that the revenue generated by selling of cortan commodity is given by function R(x)=80x-0.2x^ 2 , where he maximum reveremany should be manufactured to obtain this maximum units .Let's find out:
We have following function as [tex]R(x)=80x-0.2x^ 2[/tex] . Let's differentiate this and equate it to zero to find value of x for which the function is maximum!
⇒ [tex]R(x)=80x-0.2x^ 2[/tex]
⇒ [tex]\frac{d(R(x))}{dx}=\frac{d(80x-0.2x^ 2)}{dx}[/tex]
⇒ [tex]\frac{d(R(x))}{dx}=\frac{d(80x)}{dx}-\frac{d(0.2x^ 2)}{dx}}[/tex]
⇒ [tex]0=80-2x(0.2)[/tex]
⇒ [tex]\frac{80}{0.4}=x[/tex]
⇒ [tex]x=200[/tex]
Now , Value of function at x=200 is :
⇒ [tex]R(200)=80(200)-0.2(200)^ 2[/tex]
⇒ [tex]R(200)=16000-8000[/tex]
⇒ [tex]R(200)=8000[/tex]
Therefore , The maximum units is 200 and , Total revenue is $8,000
