# Max length possible

I have a cabinet that has 15" door. I can use $3$ baskets of $15 \times 15$ in the cabinet. I will like to know if there is any possibility that $15 \times 20$ or bigger basket can fit in this cabinet.

The problem is, it will not turn inside the cabinet if it is too large. Is there any mathematical way to find out the maximum length possible in this case?

• It can be proved via calculus that for a rectange with fixed perimeter but variable length and breadth, the area would be maximum when the rectangle is a square. Nov 10, 2013 at 4:25
• @GTXOC: I think the question is how to manoeuvre a bigger basket into the cupboard. Nov 10, 2013 at 4:26
• It is possible to fit 10" * 20" ? Nov 10, 2013 at 4:30
• It's really unclear what the restrictions are for solutions to your problem. Judging by the image, the cabinet is already constructed (maybe? IDK), is rectangular, has base dimensions $45" \times 15"$, and has a single door at one end of one of its $45"$-long sides that closes at the corner of the cabinet. (ct'd)
– Dan
Nov 10, 2013 at 4:31
• Are you asking what the side lengths are of the largest basket that can fit in this cupboard? If so, what is a "basket" for the purposes of this problem? Are we to assume that the basket is rectangular? If so, copper.hat has already pointed out in his answer that the basket must be at most $15"\times 15"$.
– Dan
Nov 10, 2013 at 4:33

You can get a semicircular basket of radius 15" into the cabinet, by sliding it in lengthwise and then rotating it about the mid-point of its diameter. This has a base area of $112.5 \pi \approx 353$ square inches, significantly more than $225$.

I realise this might not meet your requirements, but it gives the problem a new slant.

For any $\epsilon>0$, you cannot fit a basket of size $15 \times (15+\epsilon)$ into the cupboard.

It is clear that any basket that can fit inside a $15 \times 15$ square can fit straight in.

So, taking the length $l$ to be the largest dimension, we need only consider $l > 15$. (It is clear that the corresponding width must satisfy $w<15$.)

To simplify life, I am going to presume the cupboard is infinity wide. The same analysis as follows can deal with finite with cupboards, but the analysis below is complicated enough as is.

It should be clear that if we can get a basket with $l >15$ inside, we can always get it inside with the following 'trajectory': We start with the long edge vertical and the basket as far up and right as possible (that is, the right and top edges of the basket are flush against the right and top sides of the cupboard). Now rotate the basket anti-clockwise while keeping the top right edge of the basket against the top of the cupboard, with the basket as far right as possible (that is, either the the bottom right corner of the basket touches the right wall of the cupboard, or the right edge of the basket touches the bottom right corner of the cupboard.

It should be clear that each configuration above can be completely described by an angle $\theta \in [0,2 \pi]$. Given $l, \theta$, we can compute the maximum width $\bar{w}(\theta)$ that can exist for that configuration. Then a basket of length $l>15$ can be fit into the cupboard iff $w \le \bar{w}(\theta)$ for all $\theta \in [0,\frac{\pi}{2}]$.

Now for the gory details.

We need to consider two configurations.

The first is when $l \cos \theta \ge 15$.

Then we see that $\bar{w}(\theta) = 15 \cos \theta$.

The second is when $l \cos \theta < 15$.

Then we see $\frac{15-l \cos \theta}{\sin \theta}\cos \theta + \frac{1}{\cos \theta} (\bar{w}(\theta) -\frac{15-l \cos \theta}{\sin \theta} ) = 15$, which simplifies to $\bar{w}(\theta) = 15 \cos \theta + (15-l \cos \theta)\sin \theta$.

Combining gives $\bar{w}(\theta) = 15 \cos \theta + \max(0,(15-l \cos \theta)\sin \theta)$, for $\theta \in [0, \frac{\pi}{2}]$, $l > 15$.

To finish, I need to compute $w^* = \min_{\theta \in [0, \frac{\pi}{2}]} \bar{w}(\theta)$.

Let $w_1(\theta) = 15 \cos \theta$, $w_2(\theta) = 15 \cos \theta + (15-l \cos \theta)\sin \theta$, and note that $\bar{w}(\theta) = \max(w_1(\theta), w_2(\theta))$. If we let $\theta_0 = \arccos \frac{15}{l}$, we note that $\theta_0 \in (0, \frac{\pi}{2})$ and we see that $\bar{w}(\theta) = w_1(\theta)$ for $\theta \in [0,\theta_0]$ and $\bar{w}(\theta) = w_2(\theta)$ for $\theta \in [\theta_0, \frac{\pi}{2}]$.

Furthermore, $\bar{w}$ is continuous and is differentiable for $\theta \ne \theta_0$. We have $\bar{w}'(0) < 0$, $\bar{w}(0) = \bar{w}(\frac{\pi}{2}) = 15$, hence $\bar{w}$ has a minimum in $(0, \frac{\pi}{2})$.

Note that $w_1'(\theta) < 0$ for all $\theta \in (0, \frac{\pi}{2})$, and so the minimum must occur in $[\theta_0, \frac{\pi}{2})$.

We note that $w_2(0) = w_2(\frac{\pi}{2}) = 15$, and $w_2'(\theta) = (\sin \theta - \cos \theta) (l(\sin \theta + \cos \theta) -15)$. This gives $w_2'(0) = 15-l <0$, and the only zero of $w_2'$ in $[0,\frac{\pi}{2}]$ is at $\theta = \frac{\pi}{4}$. Hence $w_2$ has a strict minimum at $\theta = \frac{\pi}{4}$.

Consequently, if $\theta_0 \le \frac{\pi}{4}$ then $\bar{w}$ is minimized at $\frac{\pi}{4}$ and $w^*=\bar{w}(\frac{\pi}{4}) = 12 \sqrt{2}-\frac{l}{2}$, and if $\theta_0 > \frac{\pi}{4}$ then $\bar{w}$ is minimized at $\theta_0$, and $w^*=\bar{w}(\theta_0) = \frac{15^2}{l}$.

Translating this into a function of $l$, we see that $\theta_0 \le \frac{\pi}{4}$ iff $l \le 15 \sqrt{2}$, and so we obtain an expression for the maximum width corresponding to a given length $l$:

$$w_{\max}(l) = \begin{cases} 15,& 0 \le l \le 15 \\ 15 \sqrt{2}-\frac{l}{2},& 15 < l \le 15 \sqrt{2} \\ \frac{15^2}{l}, & 15 \sqrt{2} < l \end{cases}$$

And a plot of $w_{\max}$:

As an aside, note that $w_{\max}(20) \ge 11$, so a $20 \times 10$ basket will fit.

As a further aside (triggered by TonyK's answer), it is straightforward to show that $w_{\max} \le \frac{225}{l}$, and hence the area of the rectangle is bounded above by $225$ (and is attained at $l=15$ and for $l \ge 15 \sqrt{2}$).

I can see two places where the fit might be tightest: at $45^\circ$ if you turn it to fit it in; or else when the corner hits the far side if you slide it straight in.
If you are turning it, you need $a+\frac{b}{2}\le 15\sqrt{2}$.
If it slides straight in, at an angle $\theta$, then $\frac{a}{15}=\cos\theta=\frac{15}{b}$, so $15^2\ge ab$. I think, in this case, $\theta$ is at least $45^\circ$, so $a\le b/2$.

If you turn it, I think the tightest point is at $45^\circ$. Let the right-hand third of the cabinet be $ABCD$, with the hinge of the door at $A$, and the back of the cabinet $CD$. At the tightest point, your rectangle $PQRS$ has $P$ on $CD$, $Q$ on $BC$ and $PC=QC$. The midpoint of $RS$ is at $A$. Let $T$ be the midpoint of $PQ$. The diagonal $AC = AT+TC = a + PT = \frac{a+b}{2}\le 15\sqrt{2}$

If you don't turn it, but slide it in from the side, you don't worry about $45^\circ$. The tightest point is now when $P$ reaches the back wall, $Q$ has just reached $B$, and $RS$ is jammed against the hinge $A$. The rectangle is at an angle $\theta$, where $\cos\theta=\frac{a}{15}=\frac{15}{b}$, so $ab=15^2$ or smaller. But, as I said, $\theta$ must be more than $45^\circ$, so $\frac{a}{b}=\cos^2\theta<\frac{1}{2}$.

• Can you explain? I have no maths background. I thought my problem can be solved by mathematicians :) Nov 10, 2013 at 6:00