A manufacturer produces bolts of a fabric with a fixed width. The quantity q of this fabric (measured in yards) that is sold is a function of the selling price p (in dollars per yard), so we can write q = f(p). Then the total revenue earned with selling price p is R(p) = pf(p). A. What does it mean to say that f(20)= 10,000 and f firstderivative (20)= -350?
B. Assuming the values in part a, find R first derivative (20).

(A) the selling price is $20 per yards, and the expected yards to be sold is 10,000 yards

the derivative f'(20) is negative, which means the fabric producing company will sell 350 fewer yards when selling price is $20 per yard

(B) = R'(20) = $3000

∴the company will get extra $3000 revenue when selling price is $20 per yard

Step-by-step explanation:

A. given that

f(20)= 10,000

f'(20)= -350(first derivative)

the selling price is $20 per yards, and the expected yards to be sold is 10,000 yards

the derivative f'(20) is negative, which means the higher the price, it wil reduce the number of yards to be sold making it 350 fewer yards

(B) R(p) = p f(p)

f(20)= 10,000

f'(20)= -350(first derivative)

R(p) = p f(p)

differentiate with respect to p, using product rule

R'(p) = p f' (p) + f(p) (first derivative)

where p = 20

R'(20) = 20 f' (20) + f(20)

R'(20) = 20(-350) + 10,000

R'(20) = -7000 + 10,000

R'(20) = $3000

∴ the revenue is increasing by $3000 for every selling sold yard and increase in price per yard

x-3y=4

step-by-step e