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

Category: QuestionsWhat will be the maximum value of the function 2x^3 + 3x^2 – 36x + 10?
Editor">Editor Staff asked 11 months ago

What will be the maximum value of the function 2x^3 + 3x^2 – 36x + 10?
 
(a) 71
 
(b) 81
 
(c) 91
 
(d) 0
 
The question was asked at a job interview.
 
The origin of the question is Calculus Application in section Application of Calculus of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

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