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ruslelena [56]
3 years ago
10

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.
Mathematics
1 answer:
wariber [46]3 years ago
8 0

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.

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Answer:

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