1. ## optimization questions

I'm stuck on 2 questions here

1) An animal breeder wishes to create 5 adjacent rectangular pens, each with an area of 2400 sq meters. To ensure that the pens are large enough for grazing, the minimum for either dimensions must be 10m. Find the dimensions for the pens in order to keep the amount of fencing to a minimum.
So far i got these equations 5xy=12000
xy=2400
and i'm stuck with what to do next.

2) A train leaves the station at 10:00 and travels due north at a speed of 100km/h. Another train has been heading due west at 120km/h and reaches the same station at 11:00. At what time were the 2 trains closests?

2. Hello, imthatgirl!

2) A train leaves the station at 10:00 and travels due north at 100km/h.
Another train has been heading due west at 120km/h and reaches the station at 11:00.
At what time were the 2 trains closest?

Train #1 leaving the station $\displaystyle S$ at 100 km/hr.
In $\displaystyle t$ hours, it is at $\displaystyle A\!:\;SA = 100t$

Train #2 heads west at 120 km/hr and reaches $\displaystyle S$ at 11:00.
. . Hence, at 10:00, it was at $\displaystyle P$, 120 km east of $\displaystyle S.$

In $\displaystyle t$ hours, it is at $\displaystyle B\!:\;PB = 120t\quad\Rightarrow\quad SB = 120-120t$
Code:
    A *
| *
|   *
100t |     *
|       *
|         *
* - - - - - * - - - - - *
S 120-120t  B    120t   P
: - - - -  120  - - - - :

The distance is: . $\displaystyle d \;=\;AB^2\;=\;(100t)^2 + (120-120t)^2$

Hence: .$\displaystyle d \;=\;24,400t^2 - 28,800t + 14,400$

Then: .$\displaystyle d\,' \;=\;48,800t - 28,800 \;=\;0 \quad\Rightarrow\quad t \:=\:\frac{36}{61}\text{ hour} \;\approx\;35\text{ minutes}$

Therefore, the trains were the closest at about $\displaystyle \boxed{10:35}$