What are the radius, length and volume of the largest cylindrical package that may be sent using a parcel delivery service that will deliver a package only if the length plus the girth (distance around) does not exceed 108 inches?

koraby2bc
2022-09-23
Answered

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Hayden Espinoza

Answered 2022-09-24
Author has **7** answers

First you set up your variables and formulae:

Let r=radius and l=length

Then girth $G=2\pi r$

Volume $V=\pi {r}^{2}l$

We know that $G+l=2\pi r+l\le 108$ as per regulation.

We will take =108 to get the maximum and 'lose' the l

$l=108-2\pi r$

We can now express the volume in r only:

$V=\pi {r}^{2}\cdot l=\pi {r}^{2}\cdot (108-2\pi r)=108\pi {r}^{2}-2{\pi}^{2}{r}^{3}$

To maximise we set the derivative to 0

$V\prime =216\pi r-6{\pi}^{2}{r}^{2}=6\pi r\cdot (36-\pi r)=0\to$

$36-\pi r=0\to r=36/\pi \approx 11.46$

Substituting:

$G=2\pi r=72$

$l=108-72=36$

$V=\pi \cdot {11.46}^{2}\cdot 36=14853$

Let r=radius and l=length

Then girth $G=2\pi r$

Volume $V=\pi {r}^{2}l$

We know that $G+l=2\pi r+l\le 108$ as per regulation.

We will take =108 to get the maximum and 'lose' the l

$l=108-2\pi r$

We can now express the volume in r only:

$V=\pi {r}^{2}\cdot l=\pi {r}^{2}\cdot (108-2\pi r)=108\pi {r}^{2}-2{\pi}^{2}{r}^{3}$

To maximise we set the derivative to 0

$V\prime =216\pi r-6{\pi}^{2}{r}^{2}=6\pi r\cdot (36-\pi r)=0\to$

$36-\pi r=0\to r=36/\pi \approx 11.46$

Substituting:

$G=2\pi r=72$

$l=108-72=36$

$V=\pi \cdot {11.46}^{2}\cdot 36=14853$

asked 2022-09-19

I have a knapsack problem

$\begin{array}{rl}& \underset{x\in \{0,1{\}}^{n}}{max}\sum _{i=1}^{n}{v}_{i}{x}_{i}\\ & \text{s.t.}\sum _{i=1}^{n}{w}_{i}{x}_{i}\le c.\end{array}$

The Lagrangian relaxation is as follows

$\begin{array}{r}\underset{\lambda \ge 0}{min}\underset{x\in \{0,1{\}}^{n}}{max}\sum _{i=1}^{n}{v}_{i}{x}_{i}-\lambda (\sum _{i=1}^{n}{w}_{i}{x}_{i}-c).\end{array}$

Suppose I solved the relaxed problem and got an optimal ${x}_{lag}$ s.t. $f({x}^{\ast})<f({x}_{lag})$ where ${x}^{\ast}$ is the optimal solution of the original problem and $f$ is the objective function. Even though ${x}_{lag}$ gives a strict bound, is it consideblack to be a good approximate solution?

Is it true that the relaxation can be solved in polynomial time since the dual problem is convex in $\lambda $ and the maximization part with fixed $\lambda $ is just activating ${x}_{i}$ associated with the largest term $({v}_{i}-\lambda {w}_{i})$?

$\begin{array}{rl}& \underset{x\in \{0,1{\}}^{n}}{max}\sum _{i=1}^{n}{v}_{i}{x}_{i}\\ & \text{s.t.}\sum _{i=1}^{n}{w}_{i}{x}_{i}\le c.\end{array}$

The Lagrangian relaxation is as follows

$\begin{array}{r}\underset{\lambda \ge 0}{min}\underset{x\in \{0,1{\}}^{n}}{max}\sum _{i=1}^{n}{v}_{i}{x}_{i}-\lambda (\sum _{i=1}^{n}{w}_{i}{x}_{i}-c).\end{array}$

Suppose I solved the relaxed problem and got an optimal ${x}_{lag}$ s.t. $f({x}^{\ast})<f({x}_{lag})$ where ${x}^{\ast}$ is the optimal solution of the original problem and $f$ is the objective function. Even though ${x}_{lag}$ gives a strict bound, is it consideblack to be a good approximate solution?

Is it true that the relaxation can be solved in polynomial time since the dual problem is convex in $\lambda $ and the maximization part with fixed $\lambda $ is just activating ${x}_{i}$ associated with the largest term $({v}_{i}-\lambda {w}_{i})$?

asked 2022-08-21

How do you find the point on the the graph $y=\sqrt{x}$ which is plosest to the point (4,0)?

asked 2022-08-16

What is a tight lower bound to $\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}$ under the restrictions $\sum _{i=1}^{n}{x}_{i}=0$ and $\sum _{i=1}^{n}{x}_{i}^{2}={a}^{2}$ ?

Conjecture: due to the steeper rise of $\frac{1}{a+x}$ for negative $x$, one may keep those values as small as possible. So take $n-1$ values ${x}_{i}=-q$ and ${x}_{n}=(n-1)q$ to compensate for the first condition. The second one then gives ${a}^{2}=\sum _{i=1}^{n}{x}_{i}^{2}={q}^{2}((n-1{)}^{2}+n-1)={q}^{2}n(n-1)$. Hence,

$\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}\ge \frac{n-1}{a(1-1/\sqrt{n(n-1)})}+\frac{1}{a(1+(n-1)/\sqrt{n(n-1)})}$

should be the tight lower bound.

Conjecture: due to the steeper rise of $\frac{1}{a+x}$ for negative $x$, one may keep those values as small as possible. So take $n-1$ values ${x}_{i}=-q$ and ${x}_{n}=(n-1)q$ to compensate for the first condition. The second one then gives ${a}^{2}=\sum _{i=1}^{n}{x}_{i}^{2}={q}^{2}((n-1{)}^{2}+n-1)={q}^{2}n(n-1)$. Hence,

$\sum _{i=1}^{n}\frac{1}{a+{x}_{i}}\ge \frac{n-1}{a(1-1/\sqrt{n(n-1)})}+\frac{1}{a(1+(n-1)/\sqrt{n(n-1)})}$

should be the tight lower bound.

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