Determine the maximum possible volume, excluding the volume of the legs. A table is to be constructed by gluing together 68 cubes of dimension $1\times 1\times 1$. All four legs and the rectangular top will be formed by the cubes. The four legs must be the same length and must be one cube thick, and the top is one cube thick, as well. What is the maximum possible volume of the space between the table's top and the floor, excluding the space taken up by the table's legs?
My solution:
Let $a$ be the area of the top and $c$ be the length of a leg. 
$$
\begin{align}
a + 4c &= 68  \cr
a &= 68 - 4c
\end{align}
$$
We try to find the maximum value of ac, and then subtract 4c:
$$ac = (68 - 4c)c = -4c^2 + 68c$$
Completing the square gets $-4(c - 17/2)^2 + 289$.
The lowest value of $ac$ is when $c= 17/2$, and the lowest value is $289$. Then we subtract $289 - 4c = 289 - 34 = 255$.
Therefore, the maximum volume when excluding the volume of the legs is $255$.
However, the actual answer is $256$ as the maximum volume when excluding the legs. Where did I go wrong? 
 A: You want to maximise the volume excluding the legs, which is $ac-4c$. You maximised $ac$ instead.
We have that $ac-4c=-4c^2+64c$. 
Now complete the square:
$$ -4c^2+64c=256-(16-2c)^2$$
The maximum of $256$ is at $c=8$.
A: Noting that subtracting $68-4=64$, we can see that, if the legs are only $1$ cube high, that leaves $8\times 8\times 1=64$ for the table top. The space under the table (minus the legs) is $64-(4\times 1\times 1)=60$ so the volume from the table top to the floor is $2*64-4=124$.
If we let the legs be $4\times 2=8$, the table top is $5\times 12=60$, volume: $3*60-8=172$.
If we let the legs be $4\times 3=12$, the top is $68-12=54$ volume: $4\times 54-12=204$
If we let the legs be $4\times 4=16$, the top is $68-16=52$, volume: $5\times 52-16=244$
If we let the legs be $4\times 5=20$, the top is $68-20=48$, volume: $6\times 48-20=268$
If we let the legs be $4\times 6=24$, the top is $68-24=44,$ volume: $7\times 44-24=284$
If we let the legs be $4\times 7=28$, the top is $68-28=40,$ volume: $8\times 40-28=292$
If we let the legs be $4\times 8=32$, the top is $68-32=36,$ volume: $9\times 36-32=292$
If we let the legs be $4\times 9=36$, the top is $68-36=32,$ volume: $10\times 32-36=284$
If we let the legs be $4\times 10=40$, the top is $68-40=28,$ volume: $11\times 28-40=268$
It appears $292$ is the maximum volume with outer dimensions of $8\times 5\times 8$ or $9\times 6\times 6$
A: Given 68 cubes, the table top is $68-4h$ where $h$ is the height of the legs.
Under the table we have $(h)(68-4h)=68h-4h^2$ and the total volume (sans legs) is $top+under-legs=(68-4h)+(68h-4h^2)-(4h)=68+60h-4h^2$.
For a continuous function, the max volume can be found by where the increasing/decreasing slope of the given function is zero.
The first derivative $\frac{d}{dh}(68+60h-4h^2)=-8h+60\implies 60-8h=0\implies 8h=60$
This tells us that the max leg height is $\frac{60}{8}=7.5$ If fractional heights were allowed the max volume would be $68+60h-4h^2=68+(60\times7.5)-(4\times7.5^2)=68+450-225=293$ but we are are restricted to the nearest integers. Soooo
$$68+60h-4h^2=68+(60\times7)-(4\times49)=68+420-196=292$$
and
$$68+60h-4h^2=68+(60\times8)-(4\times64)=68+480-256=292$$
