How to find the maximum volume of a box with inversely proportional sides This problem is from my brother's calculus class, so it is slightly over my head (I'm in pre-calc), but I am curious how to go about solving this problem.
Basically there is a rectangle with sides of $15$ units and $9$ units in length, and four squares with sides that have a length of $x$ are cut out of each corner.
If you fold up each of the sides so that it makes a 3-dimensional shape, how would you figure out the maximum volume of the shape?
Here's what I have so far:
$V = (15-2x)(9-2x)x$
$V = 4x^3 - 48x^2 + 135x$
That's about as far as I've gotten. Can anybody help?
 A: Since the derivative is the slope of the curve, the point is a maximum or a minimum
when derivative equals zero
You have,
$$V = 4x^3 - 48x^2 + 135x\\
\dfrac{dV}{dx}=12x^2-96x+135\\
\dfrac{dV}{dx}=0 \Rightarrow 12x^2-96x+135 =0$$
Solving this, we get $$x\approx6.2,1.8$$
Substituting these values on $V$,(or by excluding $6.2$, since one of the sides is $(9-2x)$) we can find that maximum value of $V$ will be obtained when $x=1.8$
Therefore, Maximum Volume $=110$
This can be verified using the graph
A: The extreme values of "nice" functions (such as polynomials, which is what $V(x)$ is) occur either at boundary values or at points where derivative is zero.
Here, first get the "boundaries" of $x$, for example, $x$ cannot be negative so lower boundary is zero; also it cannot exceed the box's dimension. Find out $V$ at such points.
Then, find out $V$ at points where $V'(x)=0$
The maximum of all these $V$s is the requisite value.
A: Continuing from neo's approximate root if we take $x\approx 6.2$ then length of one side of the box would be  $ 9-2\times 6.2 <0,~$  which is discarded.
Next taking second root for $4 x^2-32x+45 =0,~ x\approx 1.820551 ;$
Volume $ = 1.820551 (9-2 \times 1.820551)(15-2 \times 1.820551)\approx 110.81908 ;$
Sign of next (second) derivative $ 8(x-4) $ is negative hence volume is confirmed maximum.
