Calculus

Application of Derivatives

Easy

applied

The revenue function is given by $R(x)=150x-2x^2-x^3$ , where $x$ is the number of units demanded, then the marginal revenue when 5 units are sold, is :

₹55

₹50

₹45

₹60

✅ Correct Option: 1

Marginal revenue is $R'(x) = 150 - 4x - 3x^2$ .

At $x = 5$ : $R'(5) = 150 - 20 - 75 = 55$ .

So the marginal revenue is ₹55.

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Match List-I with List-II

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Consider the function f(x) = 2x³ - 21x² + 36x + 80, x∈[0, 6]. Then

List-I	List-II
(A) one of its critical points is at x =	(I) -28
(B) Its absolute maximum value is	(II) -42
(C) Its absolute minimum value is	(III) 97
(D) Its second derivative at x = 0 is	(IV) 6

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The revenue function is given by R(x)=150x−2x2−x3R(x)=150x-2x^2-x^3R(x)=150x−2x2−x3, where xxx is the number of units demanded, then the marginal revenue when 5 units are sold, is :