A manufacturer 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.

Question

27-11-2019
Mathematics

contestada

A manufacturer 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.

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ChiKesselman ChiKesselman · Answer 1 · 2019-11-27T07:10:08+01:00

Answer:

[tex]R'(20) = 2000[/tex]

Step-by-step explanation:

We are given the following in the question:

Quantity, q

Selling price in dollars per yard, p

[tex]q=f(p)[/tex]

Total revenue earned =

[tex]R(p)=pf(p)[/tex]

f(20)=13000

This means that 13000 yards of fabric is sold when the selling price is 20 dollars per yard.

f′(20)=−550

This means that increasing the selling price by 1 dollar per yards there is a decrease in fabric sales by 550.

We have to find R'(20)

Differentiating the above expression, we have,

[tex]R'(p) = \displaystyle\frac{d(R(p))}{dp} = \frac{d(pf(p))}{dp} = f(p) + pf'(p)[/tex]

Putting the values, we get,

[tex]R'(p) = f(p) + pf'(p)\\\\R'(20) = f(20) + 20(f'(20))\\\\R'(20) = 13000 + 20(-550) = 2000[/tex]

AFOKE88 AFOKE88 · Answer 2 · 2021-10-22T16:28:19+02:00

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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