What will be the maximum value of the function 2x^3 + 3x^2 – 36x + 10?

(a) 71

(b) 81

(c) 91

(d) 0

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The origin of the question is Calculus Application in section Application of Calculus of Mathematics – Class 12

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Right choice is (c) 91

The explanation is: Let y = 2x^3 + 3x^2 – 36x + 10 ……….(1)

Differentiating both sides of (1) with respect to x we get,

dy/dx = 6x^2 + 6x – 36

And d^2y/dx^2 = 12x + 6

For maxima or minima value of y, we have,

dy/dx = 0

Or 6x^2 + 6x – 36 = 0

Or x^2 + x – 6 = 0

Or (x + 3)(x – 2) = 0

Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2

Now, d^2y/dx^2 = 12x + 6 = 12(-3) + 6 = -30 < 0

Putting x = -3 in (1) we get its maximum value as,

2x^3 + 3x^2 – 36x + 10 = 2(-3)^3 + 3(-3)^2 – 36(-3) + 10

= 91