The maximum number of calling minutes you can use for $50 is 210 minutes.
To solve this, we have the function cost C(x) that depends on the amount of acalling munutes (x)
We want this cost to be $50 or lower. This means:
[tex]\begin{gathered} CostFunction\colon C(x)=20+0.2(x-60) \\ Maximum\text{ value of 50:}C(x)\le50 \end{gathered}[/tex]Then we can create an inequality:
[tex]50\ge20+0.2(x-60)[/tex]And now we can solve for x:
[tex]\begin{gathered} 50\ge20+0.2(x-60) \\ \frac{50-20}{0.2}\ge x-60 \\ 150+60\ge x \\ x\le210\text{ minutes} \end{gathered}[/tex]Thus, with $50 we can talk up to 210 minutes.
To be sure of the result, let's plug x = 210 in the function and it should give us a cost of C(210) = 50:
[tex]\begin{gathered} x=210\Rightarrow C(210)=20+0.2(210-60) \\ C(210)=20+0.2\cdot150 \\ C(210)=20+30=50 \end{gathered}[/tex]This confirms the result.