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Palpa's Company manufactures bookbags, when the bookbags price is $75 the demand is 25; when the price is $20 the demand is 100 bookbags. But when the price is $50 the supply is 100 and for $25 the supply is 20.a. What is the demand function? b. What is the supply function c. What are the equilibrium price and quantity?

Sagot :

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]