# A fur dealer find that when coats sell for \$3200, monthly sales are 70 coats, when the price increases to \$3500 the demand is for 20 coats. Assume that the demand equation is linear. a) Find the...

A fur dealer find that when coats sell for \$3200, monthly sales are 70 coats, when the price increases to \$3500 the demand is for 20 coats. Assume that the demand equation is linear.

a) Find the demand and revenue equations ( interms of x, the number of coats sold monthly)

the demand equation is p=

and the revenue equation is R(x)=

Knowing that the demand quation is linear means that it will follow the form:

y=mx+b, where in this case y=p

We can use the given data to solve for m, the slope:

`m=(p_2-p_1)/(x_2-x_1)=(3500-3200)/(20-70)=-6`

Therefore, for every 6 dollar increase in price, one less coat is sold.

Substituting in one of the known points, we cna solve for b, the y-intercept:

`3500=-6(20)+b -gt b=3500+120=3620`

Therefore, at a cost of \$3620, no coats will be sold.

Therefore the demand equation is:

`p(x)=-6x+3620`

The equation for revenue will be equal to the price per coat multiplied by the number of coats sold:

`R(x)=p(x)x=-6x^2+3620x`