Part a.
The break-even point is where the cost equals to the revenue, that is,
![C(x)=R(x)](https://img.qammunity.org/2023/formulas/mathematics/high-school/wcn0vkhz6f3wfrmex35vd0xkobhgnjxbga.png)
Then, by substituting the given information, we get
![12x+29=25x](https://img.qammunity.org/2023/formulas/mathematics/high-school/clgp8dtawna4pg9058ua3cmr14wmn4brs7.png)
So, by subtracting 12x to both sides, it yields,
![\begin{gathered} 29=13x \\ or\text{ equivalently, } \\ 13x=29 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/thb00s3kzx50nlel15tpql81n26tzty8wy.png)
Then, x is given by
![x=(29)/(13)=2.2307](https://img.qammunity.org/2023/formulas/mathematics/high-school/txl4lfbho6cej9rz5ul6ksfvvdujie717w.png)
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,
![P(x)=R(x)-C(x)](https://img.qammunity.org/2023/formulas/mathematics/high-school/3r9w44q173wp0f0kgyb63izfoqnf6o9s8j.png)
So we have
![P(x)=25x-(12x+29)](https://img.qammunity.org/2023/formulas/mathematics/high-school/67b03yhi3pz5g0lse3q0uhphyu02b7t4fp.png)
which gives
![P(x)=13x-29](https://img.qammunity.org/2023/formulas/mathematics/high-school/vv89g95k9an6jprqq2vmkpp4za9u61efxr.png)
By substituting x=25o into this result ,we have
![\begin{gathered} P(250)=13(250)-29 \\ P(250)=3250-29 \\ P(250)=3221 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/40r7cpr9r2uz1irmpubnjlpn9l7k88k5c6.png)
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,
![130=13x-29](https://img.qammunity.org/2023/formulas/mathematics/high-school/peswfozds5m3q6mwg9omujz93u7y3gdri4.png)
So, by adding 29 to both sides, we have
![\begin{gathered} 159=13x \\ or\text{ equivalently, } \\ 13x=159 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/8ip5zww08zp3t0bfy2d8ow0lrhooop6l7c.png)
Therefore, we have
![\begin{gathered} x=(159)/(13) \\ x=12.23 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/j3wl29nxq8h62d26vsv1l7pru4nddctqlt.png)
Therefore, by rounding to the nearest whole number, the number of medals to produce a profit of $130 is 12 medals.