Answer

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**Hint:**

1) Revenue = Price per unit × Number of units demanded.

2) The maximum/minimum value of a function f(x) is attained when f’(x) = 0.

3) At the points of maxima, f’’(x) is < 0 and at the points of minima, f’’(x) > 0.

4) Differentiation: $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ .

**Complete step by step solution:**

Given that Number of units = x and price per unit = p.

(i) Using the formula, Revenue = Price per unit × Number of units demanded, we will get:

\[\text{Revenue}=x\times p=\left( \dfrac{24-2p}{3} \right)\times p=8p-\dfrac{2{{p}^{2}}}{3}\] , which is the required revenue function in terms of p.

(ii) Since, the revenue function is in terms of p, let us differentiate it with respect to p and equate it to 0 to find the points of maxima/minima.

$\dfrac{d}{dp}\left( 8p-\dfrac{2{{p}^{2}}}{3} \right)=0$

On differentiating, we get

⇒ $8-\dfrac{2\times 2p}{3}=0$

After taking LCM and simplify, we get

⇒ $\dfrac{4p}{3}=8$

⇒ p = 6.

Also, $\dfrac{{{d}^{2}}}{d{{p}^{2}}}\left( 8p-\dfrac{2{{p}^{2}}}{3} \right)=\dfrac{d}{dp}\left[ \dfrac{d}{dp}\left( 8p-\dfrac{2{{p}^{2}}}{3} \right) \right]=\dfrac{d}{dp}\left( \dfrac{8-4p}{3} \right)=\dfrac{-4}{3}$ .

Since, the second derivative is < 0, the value of the revenue function at p = 6 is maximum.

And, using the given relation, $x=\dfrac{24-2p}{3}=\dfrac{24-2\times 6}{3}=\dfrac{12}{3}=4$ .

The maximum value of the revenue is obtained for x = 4 units and price p = 6 per unit.

**Note:**

1) Demand function shows the relationship between quantity demanded for a particular commodity and the factors that are influencing it (like price of the commodity, price of related goods, income of consumer etc).

2) The maximum/minimum value of a quadratic function can also be found by the method of completion of squares or by using the fact that the maximum/minimum value of a quadratic function is obtained at the mean of its roots.

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