The marginal price StartFraction dp Over dx EndFraction at x units of demand per week is proportional to the price p. There is no weekly demand at a price of $100 per unit [p (0 )equals 100 ], and there is a weekly demand of 6 units at the price of $53.89 per unit [p (6 )equals 53.89 ]. Use the given information to answer the questions.

a.) Find the price demand equation.
p(x)=_______

b.) At a demand of 10 units per week, what is the price?
p(10)=$ (round to nearest cent.)

c.) how to graph the price demand equation for 0 is less than equal to x which is less than equal to 25/

We want to understand the pattern of demand so that we can make plans and see if it's compensating the demand price, usually denoted by D(q) but also p, as p for price.

1. Let's gather our information:

[tex]\frac{dp}{dx} \rightarrow (q, p)= (0,100) \:and\:(6,53.89)[/tex]

2. Assuming it's a linear then:

[tex]y=mx+b\\D(q)=mx+b \:or\:p=mx+b[/tex]

3. Let's find the slope,

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x{1}} =\frac{53.89-100}{6-0} = \frac{-46.11}{6}=-\frac{1537}{200} =[/tex]

Let's write it in this form Point Slope

[tex](y-y_1)=m(x-x_1)[/tex]

But Instead of y, let's rewrite y as p, (for price).

[tex]p-100=\frac{-1537}{200}(x-0) \therefore\: p=\frac{-1537}{200}x+100\\\\p(x)=-\frac{1537}{200}x+100[/tex]

Plugging $10, for x into the demand function:

[tex]b) p(10)=-\frac{1537}{200}(10)+100 \therefore p(10)=23.15 \therefore \$23.15[/tex]

c) Could you elaborate it better? It was not quite clear.

Anyway, here's the graph of the function notice its behavior.