# Solving cubic equations?

• Sep 25th 2012, 02:33 PM
xoxo
Solving cubic equations?
A function that represents the volume of a cardboard box is V(x) = -0.65x^3 + 4x^2 + 3x, where x is the width of the box. Determine the width that will maximize the volume. What are the restrictions on the width?

The answer is 4.45 and domain is 0<x<6.83

My teacher didnt teach me how to find the vertex and maximum of cubic equations yet so i just factored out the x and used it as a quadratic equation and put it in vertex form. But that didn't work :( HELP!

Two soccer players start at opposite sides of an 80-m field. One runs at 4 m/s and the other runs at 5 m/s. If they run back and forth for 15 mins, how many times will they pass each other?

**The answer is 25 but I keep getting 45. I didn't use any equations to use this but I keep messing with the numbers and end up at 45.
• Sep 25th 2012, 08:05 PM
johnsomeone
Re: Solving cubic equations?
Quote:

Originally Posted by xoxo
A function that represents the volume of a cardboard box is V(x) = -0.65x^3 + 4x^2 + 3x, where x is the width of the box. Determine the width that will maximize the volume. What are the restrictions on the width?

The answer is 4.45 and domain is 0<x<6.83

You find the maximimum of a function by finding, then examining and comparing:
1. where the derivative = 0
2. where the derivative does not exist
3. the endpoints of the domain

In your case, $\displaystyle V(x) = -0.65x^3 + 4x^2 + 3x$.

For #1: Compute $\displaystyle dV/dx$ (a quadratic polynomial in x), then solve $\displaystyle dV/dx = 0$ for $\displaystyle x$.
For #2: V(x) is differentiable everywhere, so nothing for #2.
For #3: Need the endpoints of the domain of V. Since x is representing a dimension of a box, must have x>0.
Also, since V represents volume, must have V(x) > 0. This is actually what restricts the domain, and takes some work to solve.

First solve V(x) = 0, the decide on which intervals V is strickly positive. (This approach relies on the Intermediate Value Theorem.)

$\displaystyle V(x) = -0.65x^3 + 4x^2 + 3x = 0 \Rightarrow -0.65x^2 + 4x + 3 = 0$ or $\displaystyle x = 0$.

If $\displaystyle V(x) = 0$ and $\displaystyle x \ne 0$, then $\displaystyle -0.65x^2 + 4x + 3 = 0$

$\displaystyle \Rightarrow x = \frac{-(4) \pm \sqrt{ (4)^2- 4(-0.65)(3)} }{2(-0.65)}$

$\displaystyle \Rightarrow x = \frac{-4 \pm \sqrt{ 16 + 4(1.95)} }{2(-0.65)} \Rightarrow x = \frac{2 \pm \sqrt{ 4 + 1.95} }{0.65}$

$\displaystyle \Rightarrow x = \frac{2 \pm \sqrt{ 5.95} }{0.65} \Rightarrow x \in \{6.83, -0.67\}$ (when $\displaystyle x \ne 0$).

But $\displaystyle V(0)= 0$ also, so the solutions to $\displaystyle V(x) = 0$ are $\displaystyle \{6.83, -0.67, 0\}$.

Thus x is either always positive, or always negative, on the intervals $\displaystyle (-\infty, -0.67), (-0.67, 0), (0, 6.83), (6.83, \infty)$.

Either by plugging in a single test value (say x = 7, so computing V(7), or by observing that V's lead coefficent is negative), it's clear the V(x) is negative for $\displaystyle x \in (6.83, \infty)$. V(1) = -0.65 + 4 + 3 > 6 > 0, so V(x) is positive for $\displaystyle x \in (0, 6.83)$. We can stop there, because V's domain requires $\displaystyle x > 0$.

Therefore the domain of V is (0, 6.83).

But as x approaches the endpoints of V's domain, V(x) goes to 0 (Note V(0) = V(6.83) = 0, and V is continuous everywhere).
Thus the endpoints of the domain will not produce anything like a maximum.

Therefore, the maximum will be produced by solving case #1 above.
• Sep 26th 2012, 12:03 AM
MaxJasper
Re: Solving cubic equations?
Quote:

Originally Posted by xoxo
Two soccer players start at opposite sides of an 80-m field. One runs at 4 m/s and the other runs at 5 m/s. If they run back and forth for 15 mins, how many times will they pass each other? **The answer is 25 but I keep getting 45. I didn't use any equations to use this but I keep messing with the numbers and end up at 45.

Player with speed 4m/s in 15 min runs the length of the pitch 45 times.
Player with speed 5m/s in 15 min runs the length of the pitch 56 times.

Within 15 min of running, I find them passing each other 56 times at times in sec:

{8.8889, 26.667, 44.444, 62.222, 80., 97.778, 115.56, 133.33, 151.11, 160., 168.89, 186.67, 204.44, 222.22, 240., 257.78, 275.56, 293.33, 311.11, 320., 328.89, 346.67, 364.44, 382.22, 400., 417.78, 435.56, 453.33, 471.11, 480., 488.89, 506.67, 524.44, 542.22, 560., 577.78, 595.56, 613.33, 631.11, 640., 648.89, 666.67, 684.44, 702.22, 720., 737.78, 755.56, 773.33, 791.11, 800., 808.89, 826.67, 844.44, 862.22, 880., 897.78}

Here is a graph of 1st 40 sec: Large dots are passing times:

http://mathhelpforum.com/attachment....1&d=1348646548
• Sep 26th 2012, 10:35 PM
johnsomeone
Re: Solving cubic equations?
Quote:

Originally Posted by xoxo
Two soccer players start at opposite sides of an 80-m field. One runs at 4 m/s and the other runs at 5 m/s. If they run back and forth for 15 mins, how many times will they pass each other?

**The answer is 25 but I keep getting 45. I didn't use any equations to use this but I keep messing with the numbers and end up at 45.

I have an unusual way to solve this. It might seem a bit overly complicated at first, but I think it's ultimately a simple way to model and solve the problem.

Think of the track having an exact copy laying next to it, then, while holding the endpoints fixed, pull the two tracks apart until they together form a circle. That makes a circle whose circumfrence is twice the length of the track, and whose leftmost and rightmost points are the two ends of the track. Now model their running back and forth across the track as them them running in circles around the circle! That motion is very easy to describe mathematically. The complication comes because points on the upper semi-circle are actually the same as points on the lower semicircle in terms of which point on the track they represent. If considered as a circle centered at the origin, it amounts to (Cartesian) (x,y) ~ (x, -y), or with angles measured in the usual manner, (Polar) Theta ~ -Theta, or (Complex) z ~ z's complex conjugate. This complication turns out to be relatively easy to deal with, and so the benefit of this model is well worth that complication.

If the track has length L, then the circumference$\displaystyle = 2L = 2 \pi R$, so the radius is $\displaystyle R = L/\pi$ (this won't matter at all).

Running back and forth at constant speed $\displaystyle v$ means you've a constant angular velocity $\displaystyle \omega$ that produces $\displaystyle \pi$ radians in the time it takes you to run one length of the track, with is $\displaystyle L/v$. Thus

Angular velocity = $\displaystyle \omega = \frac{\pi}{L/v} = \frac{\pi v}{L}$

Note that running clockwise or counterclockwise ($\displaystyle \omega$ positve or negative) on the circle makes no difference when utlimately considered as a point on the track.

Easiest is to do this in the complex plane. There, the motion is modelled as: $\displaystyle z(t) = Re^{(\theta_0 + \omega t)i}$,

where $\displaystyle \theta_0$ depends on the initial position. The track's right end corresponds to $\displaystyle \theta_0 = 0$, and its left end to $\displaystyle \theta_0 = \pi$.

With two runners at opposite sides of the track, it looks like: $\displaystyle z_1(t) = Re^{\omega_1 t i}$ and $\displaystyle z_2(t) = Re^{(\pi + \omega_2 t)i}$.

Now the problem becomes finding all t (actually, just their count) between 0 and 15 minutes (=900 seconds) such that $\displaystyle z_1(t) = z_2(t)$ OR $\displaystyle z_1(t) = \overline{z_2(t)}$.

But $\displaystyle z_1(t) = z_2(t) \Rightarrow Re^{\omega_1 t i} = Re^{(\pi + \omega_2 t)i} \Rightarrow 1 = e^{(\pi + (\omega_2 - \omega_1) t ) i}$

$\displaystyle \Rightarrow \pi + (\omega_2 - \omega_1) t = 2 \pi n \Rightarrow (\omega_2 - \omega_1) t = (2n-1) \pi$ for some $\displaystyle n \in \mathbb{Z}$.

Similarly, $\displaystyle z_1(t) = \overline{z_2(t)} \Rightarrow Re^{\omega_1 t i} = Re^{-(\pi + \omega_2 t)i} \Rightarrow e^{(\pi + (\omega_2 + \omega_1) t ) i} = 1$

$\displaystyle \Rightarrow \pi + (\omega_2 + \omega_1) t = 2 \pi n \Rightarrow (\omega_2 + \omega_1) t = (2n-1) \pi$ for some $\displaystyle n \in \mathbb{Z}$.

Write that as:

$\displaystyle z_1(t) = z_2(t) \Rightarrow (\omega_2 - \omega_1) t = \pi k_1$ for some $\displaystyle k_1 \in \mathbb{Z}^{odd}$, and

$\displaystyle z_1(t) = \overline{z_2(t)} \Rightarrow (\omega_2 + \omega_1) t = \pi k_2$ for some $\displaystyle k_2 \in \mathbb{Z}^{odd}$, and

Now for this problem:

$\displaystyle t \in [0, 900]$ (do all times in seconds), $\displaystyle L = 80$ (do all distances in meters), $\displaystyle v_1 = 4, v_2=5$, so

$\displaystyle \omega_1 = \frac{\pi v_1}{L} = \frac{4 \pi}{80}, \ \omega_2 = \frac{\pi v_2}{L} = \frac{5 \pi}{80}$, so

$\displaystyle \omega_2 + \omega_1 = \frac{5\pi}{80} + \frac{4\pi}{80} = \frac{9\pi}{80}, \ \omega_2 - \omega_1 = \frac{5\pi}{80} - \frac{4\pi}{80} = \frac{\pi}{80}$.

Thus seeking the number of $\displaystyle t \in [0, 900]$ such that $\displaystyle (\omega_2 - \omega_1) t = \pi k_1$, or $\displaystyle (\omega_2 + \omega_1) t = \pi k_2, \ k_1, k_2 \in \mathbb{Z}^{odd}$,

meaning $\displaystyle (\frac{\pi}{80}) t = \pi k_1$, or $\displaystyle (\frac{9\pi}{80}) t = \pi k_2$, which becomes:

Find the number of $\displaystyle t \in [0, 900]$ such that $\displaystyle t = 80 k_1$, or $\displaystyle t = (\frac{80}{9})k_2$, where $\displaystyle k_1, k_2 \in \mathbb{Z}^{odd}$.

Since $\displaystyle \lfloor 900/80 \rfloor = 11$, write $\displaystyle 900 = (80)(11) + (20)$, to see the count of the 1st condition is 6 ($\displaystyle k_1 \in \{1, 3, 5, 7, 9, 11\}$).

Since $\displaystyle \lfloor 900/(80/9) \rfloor = 101$, write $\displaystyle 900 = (80/9)(101) + (20/9)$, to see the count of the 2st condition is 51 ($\displaystyle k_2 \in \{1, 3, 5, ..., 99, 101\}$).

Overlap occurs where $\displaystyle 80 k_1 = (\frac{80}{9}) k_2$, so where $\displaystyle k_2 = 9 k_1$.

Since $\displaystyle k_1 \in \{1, 3, 5, 7, 9, 11\}$ and $\displaystyle k_2 \in \{1, 3, 5, ..., 99, 101\}$, you do get overlap for ALL of the first case.

(In other words, every one of the times in these solutions: $\displaystyle t = 80k_1, \ k_1 \in \{1, 3, 5, 7, 9, 11\}$, is already counted in the other set of time solutions.)

Therefore,
the times (in seconds) when the runners cross $\displaystyle = \left\{ t = \frac{80}{9} k_2 \ | \ k_2 \in \{1, 3, 5, ..., 99, 101\} \right\}$. That's 51 times.

(Note: the overlap occurs where $\displaystyle z_1(t) = z_2(t)$ and $\displaystyle z_1(t) = \overline{z_2(t)}$, which means that $\displaystyle z_1(t) = z_2(t)$ = real = left or right end of the track.
That all the first type of occurances were contained in the second type of occurances says that, in terms of the circle, the only times they meet like this $\displaystyle z_1(t) = z_2(t)$ happens when they meet at one side of the track or the other. In terms of the circle motion, there is no meeting by "catching up", except exactly at the ends of the track. That, and the cleanform of the solution, suggest that it's a very periodic pattern. In first 80 seconds, observe how many times they cross, AND where they are on the track. Thereafter, consider how many times they cross every 160 seconds, AND where they are on the track. It should be repetitive.)