Let p be the price per unit of a cetain product, when there is a sale of x units - Maths - Continuity and Differentiability

Question

Let p be the price per unit of a cetain product, when
there is a sale of x units The total revenue function :
R(X) = (100 X) / (3X+1 ) - 4X,
I) find the marginal revenue function,i e , the rate of change
of total revenue function w r t x
ii) When x = 10, find the relative change of revenue
R, i e , Rate of change of R with
respect to x and also the percentage rate of change of
R at x = 10

Priyanka Kedia · Accepted Answer

Given, Total Revenue function,

$R (x) = \frac{100 x}{3 x + 1} - 4 x ...... (1)$

(1)

Marginal Revenue function MR(x) = Rate of change of total revenue function R'(x)

$M R (x) = R' (x) = \frac{ⅆ}{ⅆ x} R (x) M R (x) = \frac{ⅆ}{ⅆ x} (\frac{100 x}{3 x + 1} - 4 x) = \frac{(3 x + 1) \frac{ⅆ}{ⅆ x} (100 x) - 100 x \frac{ⅆ}{ⅆ x} (3 x + 1)}{{(3 x + 1)}^{2}} - \frac{ⅆ}{ⅆ x} (4 x) = \frac{(3 x + 1) 100 - 100 x (3)}{{(3 x + 1)}^{2}} - 4 = \frac{100}{{(3 x + 1)}^{2}} - 4$

$\Rightarrow M R (x) = \frac{100}{{(3 x + 1)}^{2}} - 4 ...... (2)$

(2)

$Re l a t i v e c h a n g e o f r e v e n u e R = \frac{R' (x)}{R (x)}$

Substitute value or R'(x) and R(x) from equation (2) and (1) respectively,

$\frac{R' (x)}{R} = \frac{\frac{100}{{(3 x + 1)}^{2}} - 4}{\frac{100}{(3 x + 1)} - 4 x} = \frac{100 - 4 {(3 x + 1)}^{2}}{(3 x + 1) [100 - 4 x (3 x + 1)]}$

Relative change of revenue R at (x = 10) $= {(\frac{R' (x)}{R})}_{a t x = 10}$

$= \frac{100 - 4 {(3 \times 10 + 1)}^{2}}{(3 \times 10 + 1) [100 - 4 \times 10 (3 \times 10 + 1)]} = \frac{100 - 4 \times 31^{2}}{31 (100 - 40 \times 31)} = \frac{100 - 4 \times 961}{31 (100 - 1240)} = \frac{100 - 3844}{31 \times (- 1140)} = \frac{3744}{35340} = 0.105942275$

Therefore, relative change of revenue R at (x=10) is 0.105942275

$P e r c e n t a g e r a t e o f c h a n g e o f R a t (x = 10) = 100 \times \frac{R' (x)}{R (x)} = 100 \times .105942275 = 10.594$

Therefore, percentage rate of change of R at (x=10) is 10.594