Finding the maximum value of $7t -11 - 2t^2$

Question

I am to determine the highest possible value of $p(t)$ when

$$p(t) = 7t - 11 - 2t^2$$

I try completing the square:

$$= -2\left(t^2 -\frac{7}{2}t + \frac{11}{2}\right)$$

$$= -2\left(\left(t-\frac{7}{4}\right)^2 - \frac{49}{16} + \frac{11}{2}\right)$$

$$= -2\left(\left(t-\frac{7}{4}\right)^2 + \frac{39}{16}\right)$$

But I have no idea where to go next.

Answer 1

Well the whole thing is multiplied by $-2$, meaning that it'll be largest when what's inside the bracket is smallest (since "smaller" negative numbers are in fact larger). So we want to minimize

$$(t-\frac 7 4)^2+\frac{39}{16}$$

By picking a certain $t$. The $\frac{39}{16}$ doesn't change with $t$, so it won't affect our minimization. We can throw it out. So we've reduced the problem to minimizing

$$(t-\frac 7 4)^2$$

which is simple: squares are always positive (or zero), so the smallest value this can take is zero. And it takes this value only when $t=\frac{7}{4}$. So the function takes its largest value at $t=\frac 7 4$. Plugging this into the function will give you the maximum value that the function takes.

Answer 2

Complete the square: $7t - 11 - 2t^2 = -2(t-\frac{7}4)^2-\ldots$