Answer :
Summarizing the information:
Price: $75 => Demand: 25
Price: $20 => Demand: 100
Price: $50 => Supply: 100
Price: $25 => Supply: 20
a.
To calculate the demand function, we identify at least two points of the demand line. Using the information given by the problem, these two points are:
[tex]\begin{gathered} D_1=(75,25) \\ D_2=(20,100) \end{gathered}[/tex]Given the slope formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]Where:
[tex]\begin{gathered} (x_1,y_1)=(75,25)_{} \\ (x_2,y_2)=(20,100) \end{gathered}[/tex]Then:
[tex]\begin{gathered} m=\frac{100-25}{20-75}=\frac{75}{-55} \\ \Rightarrow m=-\frac{15}{11} \end{gathered}[/tex]Now, by the slope-intercept form of the line equation (and using the point D₁):
[tex]\begin{gathered} y-25=-\frac{15}{11}(x-75) \\ y=-\frac{15}{11}x+\frac{1125}{11}+25 \\ \Rightarrow f(x)=-\frac{15}{11}x+\frac{1400}{11} \end{gathered}[/tex]And that is the demand function.
b.
We identify two points of the supply line:
[tex]\begin{gathered} S_1=(50,100) \\ S_2=(25,20) \end{gathered}[/tex]The slope of the line is:
[tex]m=\frac{20-100}{25-50}=\frac{-80}{-25}=\frac{16}{5}[/tex]Now, by the slope-point form of the line (using S₁):
[tex]\begin{gathered} y-100=\frac{16}{5}(x-50) \\ y=\frac{16}{5}x-160+100 \\ \Rightarrow g(x)=\frac{16}{5}x-60 \end{gathered}[/tex]Where g(x) is the supply function.
c.
To find the equilibrium point, we solve the equation f(x) = g(x):
[tex]\begin{gathered} -\frac{15}{11}x+\frac{1400}{11}=\frac{16}{5}x-60 \\ \frac{1400}{11}+60=\frac{16}{5}x+\frac{15}{11}x \\ \frac{2060}{11}=\frac{251}{55}x \\ \Rightarrow x=\frac{10300}{251}\approx\text{ \$}41.04 \end{gathered}[/tex]Now, we find the corresponding quantity:
[tex]\begin{gathered} f(\frac{10300}{251})=-\frac{15}{11}x+\frac{1400}{11} \\ \Rightarrow f(\frac{10300}{251})\approx71 \end{gathered}[/tex]The equilibrium point is:
[tex](\text{\$}41.04,71)[/tex]