Amanufacturer produces bolts of a fabric with a fixed width. a 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).

find r′(20), given f(20)=13000 and f′(20)=−550. also explain, what these equations mean.

A) The expression f(20) = 13000 means that;

the quantity of fabric sold was 13000 yards when the selling price was $20 per yard.

B) The expression f′(20) = −550 means that;

when the selling price in dollars is increased by $1 per yard, the quantity of fabric sold decreases by 550 yards.

C) The derivative of the revenue which is R′(20) is given as;

R′(20) = $2000

We are given;

Quantity of fabric sold; q = f(p)

Total revenue earned; R(p) = p*f(p)

where p in both cases is selling price in dollars per yard

We are told that f(20) = 13000.

This means that the quantity of fabric sold was 13000 yards when the selling price was $20 per yard.

We are told that;

f′(20) = −550

This means that when the selling price in dollars is increased by $1 per yard, the quantity of fabric sold decreases by 550 yards.

Now, since; R(p) = p*f(p), then by differentiation;

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

Thus, R′(20) is;

R′(20) = f(20) + p(f'(20))

Plugging in the relevant values gives;

R′(20) = 13000 + 20(-550)

R′(20) = $2000

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10 * 563.67 = 5636.7