Rectangle with sides $(x+y+z)$ and $(a+b+c)$ is divided into 9 smaller rectangles. Find the minimum possible area of the rectangle.
The areas of 4 of the smaller rectangles are:
$$ax=8, \quad ay=10, \quad by=5, \quad cz=12$$
So, $$P=35+az+bx+bz+cx+cy$$
I got: $$a=2b, y=\frac54x \implies P=39+ \frac32az +\frac94cx$$
How do I continue from here?
Best Answer
Start by writing $z=12/c$. You get
$$P = 39 + 18\frac ac + \frac94cx$$
Now put $x=8/a$
$$P = 39 + 18\frac ac + 18\frac ca$$
Put $a/c=t$
$$P=39+18t + \frac{18}{t}$$
You can quickly see that we need to minimise the last two terms. You can see by inspection or differentiation that it is minimum at $t=1$ and $t=-1$. Since ratio of two positive quantities cannot be negative, we take $t=a/c=1$.
So,
$$P=39+18\cdot1 +\frac{18}{1} = 39+18+18=75$$