1. ## Advanced functions finding value of t that gives maximum

I've been trying to find what time, t gives the maximum altitude for this question. I've found the answer using derivative way however i'm taking a advanced functions course and the teacher won't allow the calculus way so is there another way to find the value of t that gives the maximum altitude from this equation:

h(t) = -16t^2 + 90t + 10 000

2. Originally Posted by Devi09
I've been trying to find what time, t gives the maximum altitude for this question. I've found the answer using derivative way however i'm taking a advanced functions course and the teacher won't allow the calculus way so is there another way to find the value of t that gives the maximum altitude from this equation:

h(t) = -16t^2 + 90t + 10 000
Complete the square

$\displaystyle -16t^2+90t+10000=-16(t^2-\frac{90}{16}t+ \quad)+10000=-16\left(t^2-\frac{90}{16}+\left[ \frac{90}{32}\right]^2 \right)+10,000+16\left[ \frac{90}{32}\right]^2$

$\displaystyle =-16\left(t-\frac{90}{32} \right)^2+\frac{162025}{16}$

Now the parabola is in vertex from so you can just read the max off.

3. $h(t) = -16t^2 + 90t + 10 000$
Alternatively, you could think of the graph of this function.

It's a negative parabola. Parabolas are symmetrical along their maxima or minima.

So the t co-ordinate you require will lie along the quadratic's line of symmetry. The way to find the line of symmetry is straightforward:

Put the function equal to 0 and solve, to find the t-intercepts.

$0 = -16t^2 + 90t + 10 000$
$8t^2 - 45t - 5000 = 0$

This gives $t\approx 27.97$ or $t\approx-22.34$

Not very nice.

The line of symmetry will lie exactly half-way between them:

$\frac{27.97-22.34}{2} = 2.815$

This is the same result (approximately) that you'd get using calculus - for an exact answer, solve the quadratic exactly, leaving your answer in surd form - the surds should cancel and you get $\displaystyle\frac{90}{32}$. Not very impressive, but heck - it works.