Question

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/

Answer 1

`a) p(x)=-(1537)/(200)x+100`

`b) p(10)=-(1537)/(200)(10)+100 \therefore p(10)=23.15 \therefore \$23.15`

Explanation:

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:

`(dp)/(dx) \rightarrow (q, p)= (0,100) \:and\:(6,53.89)`

2. Assuming it's a linear then:

`y=mx+b\\D(q)=mx+b \:or\:p=mx+b`

3. Let's find the slope,

`m=\frac{y_(2)-y_(1)}{x_(2)-x{1}} =(53.89-100)/(6-0) = (-46.11)/(6)=-(1537)/(200) =`

Let's write it in this form Point Slope

`(y-y_1)=m(x-x_1)`

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

`p-100=(-1537)/(200)(x-0) \therefore\: p=(-1537)/(200)x+100\\\\p(x)=-(1537)/(200)x+100`

Plugging $10, for x into the demand function:

`b) p(10)=-(1537)/(200)(10)+100 \therefore p(10)=23.15 \therefore \$23.15`

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

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