1
$\begingroup$

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.enter image description here

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?

$\endgroup$
3
  • $\begingroup$ A couple of things to note: (1) in the question title, you're asking for the maximum (surface) area, but in the problem, you're looking at the volume. These will not necessarily have the same answer. (2) Double check your $V$ calculation; it should have $135x$. $\endgroup$
    – Théophile
    Nov 3, 2022 at 13:27
  • $\begingroup$ Yeah you're right. I'll fix the issues $\endgroup$
    – Nathan
    Nov 3, 2022 at 13:46
  • $\begingroup$ Now take a derivative and set to zero $\endgroup$
    – Ross Millikan
    Nov 3, 2022 at 14:07

3 Answers 3

Reset to default
2
$\begingroup$

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

$\endgroup$
5
  • 1
    $\begingroup$ Note that $6.2$ and $1.8$ are approximations, not exact values. There is no need to substitute $x \approx 6.2$, because we know that $x \le \frac92$ given that one of the side lengths is $9$. $\endgroup$
    – Théophile
    Nov 3, 2022 at 16:10
  • $\begingroup$ Also the statement "derivative equals zero means the the point is either a maximum or a minimum" is not true; consider, for example, $y=x^3$ at $x=0$. $\endgroup$
    – Théophile
    Nov 3, 2022 at 18:24
  • $\begingroup$ He did not take 6.2 into calculation but did not mention why.. $\endgroup$
    – Narasimham
    Nov 3, 2022 at 22:01
  • $\begingroup$ @Théophile thanks for teaching me that! $\endgroup$
    – neo
    Nov 4, 2022 at 5:49
  • $\begingroup$ @neo Glad to help. If you want to learn about some much wilder behaviour related to points where the derivative is zero, check this out: en.wikipedia.org/wiki/Cantor_function. $\endgroup$
    – Théophile
    Nov 4, 2022 at 16:56
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This is incorrect; see the comments on neo's answer. $\endgroup$
    – Théophile
    Nov 3, 2022 at 18:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .