Question

Find the maximum and minimum values of function $y=2x^3-15x^2+36x+11$

• A. $42;41$
• B. $39;38$
• C. $35;34$
• D. $52;51$

Answer 1

The correct option is B $39;38$

$y=2x^3-15x^2+36x+11$
$\frac{dy}{dx}=6x^2-30x+36$
$\frac{dy}{dx}=0=6x^2-30x+36=0$
$x^2-5x+6=0$
$x^2-2x-3x+6=0$
$(x-3)(x-2)=0$
Roots of the quadratic equation are
$x=2,3$

Now double differentiation of given equation
$\frac{d^{2}y}{d{x}^2}=12x-30$
At $x=2$, $\frac{d^{2}y}{d{x}^2}=-6$
Hence at $x=2$ there is maxima of value $y_{\text{max}}=39$
At $x=3$, $\frac{d^{2}y}{d{x}^2}=+6$
Hence at $x=3$, there is minima of value $y_{\text{min}}=38$