Part a.
The break-even point is where the cost equals to the revenue, that is,
[tex]C(x)=R(x)[/tex]
Then, by substituting the given information, we get
[tex]12x+29=25x[/tex]
So, by subtracting 12x to both sides, it yields,
[tex]\begin{gathered} 29=13x \\ or\text{ equivalently, } \\ 13x=29 \end{gathered}[/tex]
Then, x is given by
[tex]x=\frac{29}{13}=2.2307[/tex]
Then, by rounding to the nearest whole number, the break even quantity is 2 medals.
Part b.
The profit is equal to the revenue minus the cost, that is,
[tex]P(x)=R(x)-C(x)[/tex]
So we have
[tex]P(x)=25x-(12x+29)[/tex]
which gives
[tex]P(x)=13x-29[/tex]
By substituting x=25o into this result ,we have
[tex]\begin{gathered} P(250)=13(250)-29 \\ P(250)=3250-29 \\ P(250)=3221 \end{gathered}[/tex]
Therefore, the profit from 250 units is $3221.
Part c
In this case, we need to substitute P=130 into the profit function and find x, that is,
[tex]130=13x-29[/tex]
So, by adding 29 to both sides, we have
[tex]\begin{gathered} 159=13x \\ or\text{ equivalently, } \\ 13x=159 \end{gathered}[/tex]
Therefore, we have
[tex]\begin{gathered} x=\frac{159}{13} \\ x=12.23 \end{gathered}[/tex]
Therefore, by rounding to the nearest whole number, the number of medals to produce a profit of $130 is 12 medals.