Question

Find the local minimum and local maximum of y = 2x³ – 3x² – 36x + 10

Answer 1

y = 2x³ – 3x² – 36x + 10 

\(\frac{dy}{dx}\) = 6x² – 6x – 36 = 6(x² – x – 6) 

\(\frac{dy}{dx}\) = 0 gives 6(x² – x – 6) = 0 

6(x – 3) (x + 2) = 0 

x = 3 (or) x = -2 

\(\frac{d^2y}{dx^2}\) = 6(2x – 1)

Case (i): when x = 3, 

\((\frac{d^2y}{dx^2})_{x=3}\) = 6(2 x 3 – 1) 

= 6 x 5 

= 30, positive 

Since \(\frac{d^2y}{dx^2}\) is positive y is minimum when x = 3. 

The local minimum value is obtained by substituting x = 3 in y. 

Local minimum value = 2(3³) – 3(3²) – 36(3) + 10 

= 2(27) – (27) – 108 + 10 

= 27 – 98 

= -71

Case (ii): when x = -2, 

\((\frac{d^2y}{dx^2})_{x=-2}\) = 6(-2 x 2 – 1) 

= 6 x -5

= -30, negative 

Since \(\frac{d^2y}{dx^2}\) is negative, y is maximum when x = -2. 

Local maximum value = 2(-2)³ – 3(-2)² – 36(-2) + 10 

= 2(-8) – 3(4) + 72 + 10 

= -16 – 12 + 82 

= -28 + 82 

= 54