Question

The demand function for q units of a product at $p per unit is given by p(q + 3)2 = 100,000. Find the rate of change of quantity with respect to price when p = $40.

Answer 1

Answer:

-0.625 unit

Step-by-step explanation:

Given that:

$p(q+3)^2 = 100, 000$

where;

p = price

q = quantity

To find the rate of change of quantity (q) with respect to price (p) we go by the differentiation

$(q+3)^2\frac{dp}{dp} +p^2(q+3)\frac{dq}{dp}=0$

$(q+3)^2 +p^2(q+3)\frac{dq}{dp}=0$

$\frac{dq}{dp}=\frac{-(q+3)}{2p}$

when P =40

Then 4(q+3)² = 100, 000

(q+3)² = $\frac{100,000}{4}$

(q+3)² = 25,000

(q+3) = $\sqrt{2500}$

q+ 3 = 50

q = 50 -3

q = 47

NOW; $\frac{dq}{dp}=\frac{-(q+3)}{2p}$

$\frac{dq}{dp}=\frac{-(47+3)}{2*40}$

$\frac{dq}{dp}=\frac{-47-3}{2*40}$

$\frac{dq}{dp}=\frac{-50}{80}$

$\frac{dq}{dp}=-0.625$

Thus, the rate of change of quantity with respect to price when p = $40 is -0.625 unit.

Answer 2

Applying implicit differentiation, it is found that the rate of change of quantity with respect to price when p = $40 is of -0.3125 units.

The demand function is:

$p(q + 3)^2 = 100000$

When p = 40, we have that:

$p(q + 3)^2 = 100000$

$40(q + 3)^2 = 100000$

$(q + 3)^2 = \frac{100000}{40}$

$(q + 3)^2 = 2500$

$\sqrt{(q + 3)^2} = \sqrt{2500}$

$q + 3 = 25$

$q = 22$

Applying implicit differentiation, we have that

$(q + 3)^2\frac{dp}{dp} + 2p(q + 3)\frac{dq}{dp} = 0$

$2p(q + 3)\frac{dq}{dp} = -(q + 3)^2$

$\frac{dq}{dp} = -\frac{(q + 3)^2}{2p(q + 3)}$

$\frac{dq}{dp} = -\frac{q+3}{2p}$

Hence, considering that q = 22:

$\frac{dq}{dp} = -\frac{q+3}{2p} = -\frac{25}{80} = -0.3125$

The rate of change is of -0.3125 units.

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