Given the curve $y=2x^3+3x^2−36x$ on the interval $− 1 ≤ x ≤ 4$ , find the absolute maximum?

Question

I have found the first derivative which is $y′=6x^2+6x−36$.

Answer 1

First, I want to add the information that if the interval is now $[-3,4]$, then:$y' = 6(x+3)(x-2)=0\implies x=-3,2$ are critical values.

$y'' = 12x+6 = -30 < 0$ when $x = -3$, and equals to $30>0$ when $x = 2$.

Thus we have a relative maxima at $x = -3$ which is $y(-3) = 2(-3)^3+3(-3)^2 - 36(-3)= -54+27+108 = 81$.

Evaluating $y$ at both end points $x = -1,4$ yield $y(-1) = 2(-1)^3+3(-1)^2-36(-1) = -2+3+36=37$, and $y(4) = 2(4)^3+3(4)^2 - 36(4)=128+48-144=176-144= 32$.

Thus $y_{\text{max}} = 81$ occurs when $x = -3$.

If the interval is $[-1,4]$, then $-3$ is "outside" the interval. Hence its a contest between $-1$ and $4$ and the calculation above shows $y_{\text{max}} = 37$ at $x = -1$.

Answer 2

The absolute maximum of a continuous function on a closed interval occurs either on the boundaries of the interval or inside it. Let's start with the value of the function on the bounderies $$y(-1)=-2+3+36=37$$ $$y(4) = 128+48-144=32$$ Now, inside the interval the maximuma and minima will correspond to points where the derivative equals zero, since the slope the will be zero. Solving for when the derivative equal to zero. $$y'=6x^2+6x-36=0$$ We get $x = 2,-3$. Of these two only 2 is in the given interval. Let's plug it in to see if it's a max or min and if it's local or global. $$y(2) = -44$$ Thus the maximum value on the given interval occurs when $x=-1$. The Global max is $(-1,37)$.