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Convexity

The maximum v...

Question

The maximum value of $y = 2 x^{3} - 3 x^{2} - 36 x + 10$ is

A

51

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B

52

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C

53

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D

54

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Solution

The correct option is D 54
$f (x) = 2 x^{3} - 3 x^{2} - 36 x + 10$
$f^{'} (x) = 6 x^{2} - 6 x - 36$
For maxima or minima,
$f^{'} (x) = 0$
$\Rightarrow x = 3, - 2$
$f^{''} (x) = 12 x - 6$
$f^{''} (3) = 30 > 0$ and $f^{''} (- 2) = - 30 < 0$
Hence f(x) has a maximum at $x = - 2$
$f (- 2) = 54$

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Similar questions

Q. Find the maximum and minimum values of function

y = 2 x^{3} - 15 x^{2} + 36 x + 11

y = \frac{6 x^{3} - 45 x^{2} + 108 x + 2}{2 x^{3} - 15 x^{2} + 36 x + 1}

x \in (0, 10)

Maximum value of

y

y = 2 x^{3} - 21 x^{2} + 36 x - 20

(x \in (0, 8))

Then the maximum value of

y

in this period is:

Q. Find the intervals in which the following functions are increasing or decreasing.
(i) f(x) = 10 − 6x − 2x²

(ii) f(x) = x² + 2x − 5

(iii) f(x) = 6 − 9x − x²

(iv) f(x) = 2x³ − 12x² + 18x + 15

(v) f(x) = 5 + 36x + 3x² − 2x³

(vi) f(x) = 8 + 36x + 3x² − 2x³

(vii) f(x) = 5x³ − 15x² − 120x + 3

(viii) f(x) = x³ − 6x² − 36x + 2

(ix) f(x) = 2x³ − 15x² + 36x + 1

(x) f(x) = 2x³ + 9x² + 12x + 20

(xi) f(x) = 2x³ − 9x² + 12x − 5

(xii) f(x) = 6 + 12x + 3x² − 2x³

(xiii) f(x) = 2x³ − 24x + 107

(xiv) f(x) = −2x³ − 9x² − 12x + 1

(xv) f(x) = (x − 1) (x − 2)²

(xvi) f(x) = x³ − 12x² + 36x + 17

(xvii) f(x) = 2x³ − 24x + 7

(xviii)

f (x) = \frac{3}{10} x^{4} - \frac{4}{5} x^{3} - 3 x^{2} + \frac{36}{5} x + 11

(xix) f(x) = x⁴ − 4x

(xx)

f (x) = \frac{x^{4}}{4} + \frac{2}{3} x^{3} - \frac{5}{2} x^{2} - 6 x + 7

(xxi) f(x) = x⁴ − 4x³ + 4x² + 15

(xxii) f(x) = 5x³^/2 − 3x⁵^/2, x > 0

(xxiii) f(x) = x⁸ + 6x²

(xxiv) f(x) = x³ − 6x² + 9x + 15

(xxv)

f (x) = {\{x (x - 2)\}}^{2}

Q. Find the maximum and minimum values of function

y = 2 x^{3} - 15 x^{2} + 36 x + 11

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