Quadratic equation application

[paulmdrdo1]

please help me get started with these problems.

1.) It took a faster runner 10 sec longer to run a distance of 1500 ft than it took a slower runner to run a distance of 1000 ft. If the rate of the faster runner was 5ft/sec more than the slower runner, what was the rate of each?

2.) It is desired to have an open bin with a square bottom, rectangular sides, and a height of 3m. If the material for the bottom costs \$5.40 per square meter and the material for the sides costs \$2.40 per square meter, what is the volume of the bin that can be constructed for \$63 worth of material?

THANKS!

 

[MarkFL]

Let's begin with the first problem...

Let distances be measured in feet and time in seconds. Let $t$ be the time and $v$ be the speed of the slower runner. Using the relation between distance, constant speed and time:

\(\displaystyle d=vt\)

Can you construct the equations for the two runners? You will then have two equations in two unknowns.

 

[paulmdrdo1]

$vt=1000$
$(v+5)(t+10)=1500$

do you mean like this? but I expect to get a quadratic equation. how's that?

 

[MarkFL]

Since you are asked for $v$, solve the first equation for $t$, and substitute into the second equation...it looks like a quadratic in $v$ will result. :D

 

[paulmdrdo1]

my answers are

20ft/sec --slower runner
25ft/sec --faster runner

or

25ft/sec--- slower
30ft/sec--- faster

now how about the second problem?

 

[MarkFL]

We know the height $h$ is 3 m, but we don't know the length $x$ of the sides of the square bottom. We know the volume $V$ is:

\(\displaystyle V(x)=3x^2\)

The cost function is the area of the bottom times the cost in dollars per square meter of the bottom, plus the total area of the sides times the cost in dollars per square meter of the sides. Equating this to 63 will result in a quadratic in $x$.

 

[paulmdrdo1]

$2(5.40)(x^2)+4(2.40)(3)(x)=63$

$10.8x^2+28.8x-63=0$

completing the square

$x^2+\frac{8}{3}x+(\frac{4}{3})^{2}=\frac{35}{6}+(\frac{4}{3})^2$

$x = 1.42m$

$V=(1.42)^2\times 3$

$V=6.05m^3$

is my answer correct?

 

[MarkFL]

You have an open bin rather than a closed box...

 

[paulmdrdo1]

Oh yes! I just got excited and Included the part that should be open.

the answer is $V=8.33m^3$

 

[MarkFL]

I wound up with:

\(\displaystyle 3x^2+16x-35=0\)

\(\displaystyle (3x-5)(x+7)=0\)

Discarding the negative root, we are left with:

\(\displaystyle x=\frac{5}{3}\)

And so, the volume in cubic meters is:

\(\displaystyle V\left(\frac{5}{3}\right)=3\left(\frac{5}{3}\right)^2=\frac{25}{3}\)

 

[paulmdrdo1]

I wound up with:
\(\displaystyle 3x^2+16x-35=0\)

I'm just curious how did arrive at this?

 

[MarkFL]

I'm just curious how did arrive at this?

I had:

\(\displaystyle 5.4x^2+28.8x-63=0\)

Then I multiplied through by \(\displaystyle \frac{10}{18}=\frac{5}{9}\)

to get:

\(\displaystyle 3x^2+16x-35=(3x-5)(x+7)=0\)

 

[paulmdrdo1]

how did you know that multiplying the expression by that amount would give you a nice form? that's cool. please let me know.

 

[MarkFL]

how did you know that multiplying the expression by that amount would give you a nice form? that's cool. please let me know.

Multiplying by 10 gets rids of the decimal point, and then the resulting numbers have 18 as their GCD. This gives you a quadratic with the same roots and having the smallest possible integer coefficients.