Hello Cedric Cajigas,
1.) Let's let $u$ be the units filled at a rent of $r$. We wish to express $u$ as a function of $r$. We are given:
$$u(3600)=50$$
We are also told that $$\frac{\Delta u}{\Delta r}=-\frac{1}{100}$$
And so, using the point slope formula, we find:
$$u-50=-\frac{1}{100}(r-3600)$$
Hence:
$$u(r)=-\frac{r}{100}+86=\frac{8600-r}{100}$$
Now, the revenue $R$ obtained is the product of the number of units filled and the monthly rent, thus we may state:
$$R(r)=r\cdot u(r)=\frac{r(8600-r)}{100}$$
To maximize the revenue, given that it is a parabolic function opening downward, we may find the axis of symmetry midway between the roots, at:
$$r=4300$$
And so:
$$R_{\max}=R(4300)=184900$$
Thus, we have found that when the rent per unit is \$4300 per month, the monthly revenue is maximized at \$184,900.
2.) We want to set $h(t)=250$ and then solve for $t$:
$$-5t^2+100t=250$$
Divide through by -5, then add 50 to both sides to get the quadratic in standard form:
$$t^2-20t+50=0$$
Apply the quadratic formula:
$$t=\frac{20\pm\sqrt{(-20)^2-4(1)(50)}}{2(1)}=\frac{20\pm\sqrt{200}}{2}=10\pm5\sqrt{2}=5\left(2\pm\sqrt{2} \right)$$
The smaller root is for when the bullet is on its way up, and the larger root is for when the bullet is on its way back down.
3.) Let's let $v$ be the original slower speed and $t$ be the time it takes at this speed. Since the distance is the same in both cases, then we may write (using $d=vt$):
$$d=vt=(v+9)(t-2)$$
Expanding the right side, we have:
$$vt=vt-2v+9t-18$$
Add $2v-vt$ to both sides:
$$2v=9t-18$$
Multiply through by $v\ne0$:
$$2v^2=9vt-18v$$
Since $$d=vt$$ we may now arrange this in standard form as:
$$2v^2+18v-9d=0$$
Applying the quadratic formula, we find:
$$v=\frac{-18\pm\sqrt{18^2-4(2)(-9d)}}{2(2)}=\frac{3\left(-3\pm\sqrt{2d+9} \right)}{2}$$
Since we should assume that $0<v$, we take only the positive root:
$$v=\frac{3\left(\sqrt{2d+9}-3 \right)}{2}$$
