# Maximum Value on an Interval

## Question

If a toy car moves according to the equation $v(t)=2/3t^3-4t^2+6t+2$, where t represents time and v represents velocity, what is the toy car’s maximum acceleration on the interval [0, 5]?

## Solution

Acceleration is the first derivative of velocity. So if $v(t)=2/3t^3-4t^2+6t+2$ represents the toy car’s velocity, then $a(t)=2t^2-8t+6$ represents its acceleration.

### How to find the maximum?

The maximum on the interval could be a local maximum somewhere within the interval, or it could be a value at one end or the other. You can calculate all three and compare them. To get started a sketch may help you get a sense of what the curve looks like.

To make a quick sketch of $a(t)=2t^2-8t+6$, factor the trinomial to $2(t-3)(t-1)$. That tells you the x-intercepts: $(3,&space;0)$ and $(5,&space;0)$. And the 6 at the end of $2t^2-8t+6$ tells you that the y-intercept is $(0,&space;6)$. Those three points are enough for a crude sketch. I’ve extended the domain to 5 on the right, because that’s the interval that’s given. I think the y-value is higher at $x=5$ than it is at $x=0$, because 5 is farther from 2, which appears to be the vertex.

As you can see from either the sketch or the equation (the leading coefficient is positive), this parabola opens upward. Setting the acceleration function’s first derivative equal to zero will tell you where the vertex is – but as you can see that vertex is a minimum, not a maximum, so that is not the route to pursue to find the maximum.

### Where the first derivative is zero or at  the interval’s end

Since the vertex is the only turning point on the interval, the maximum value on the interval must be at one of the interval’s ends. If the sketch is accurate, the acceleration function, $a(t)=2t^2-8t+6$, should be at its maximum in the interval at $t=5$.

To confirm, you may want to calculate acceleration at several key points. To find acceleration at the left-hand end of the interval, $t=0$, is easy: Substitute 0 for t and you get $a(0)=6$.

Next, try the local minimum. To find the y-value there, you first have to find the t-value. To do that, differentiate $a(t)=2t^2-8t+6$ and get $a’(t)=4t-8$. Set that to 0 to get

$4t-8=0$

$t=2$

(Alternatively, since the acceleration function is second degree, if you represent its equation as $at^2+bt+c$, you can find its vertex by setting t equal to $-b/(2a)$. For the equation $a(t)=2t^2-8t+6$$a=2$ , $b=-8$, and $c=6$, so $-b/(2a)=2$.)

$a(2)=2(2^2)-8(2)+6=-2$

Finally, to find $a(5)$ you plug 5 into $a(t)=2t^2-8t+6$, which is a bit more work than plugging in 0, but still not a big deal. Then

$a(5)=2(5^2)-8(5)+6=16$

The results:

 t A(t) 0 6 2 -2 5 16

As the crude sketch indicated, the highest value is at $t=5$. There $a(t)=16$, and that is the answer. If you can work with a calculator and are not limited to what you can sketch by hand, you may get a better picture, like this: