Welcome to our community

Be a part of something great, join today!

Cedric Cajigas' questions at Yahoo! Answers regarding applications of quadratic equations

  • Thread starter
  • Admin
  • #1


Staff member
Feb 24, 2012
Here are the questions:

Word problems involving quadratics?

1.) The owner of an apartment building with 50 units has found that if the rent for each unit is $3,600 per month, all of the units will be filled. But one unit will become vacant for each $100 increase in the rent. What monthly rent would maximize the monthly income? What is the maximum income?

2.) If a gun is fired upward with a velocity of 100 m/s, its height after t seconds is -5t^2+100t meters. At what time will the bullet be 250 m. high?

3.) In driving to Manila, a motorist discovered that he could make the trip in the expressway in 2 hrs. less time by increasing his speed by 9 km/h. What was his original speed?

Can you please solve these? :( And if you have time, kindly explain this for me. thanks
I have posted a link there to this topic so the OP can see my work.
  • Thread starter
  • Admin
  • #2


Staff member
Feb 24, 2012
Hello Cedric Cajigas,

1.) Let's let $u$ be the units filled at a rent of $r$. We wish to express $u$ as a function of $r$. We are given:

\(\displaystyle u(3600)=50\)

We are also told that \(\displaystyle \frac{\Delta u}{\Delta r}=-\frac{1}{100}\)

And so, using the point slope formula, we find:

\(\displaystyle u-50=-\frac{1}{100}(r-3600)\)


\(\displaystyle u(r)=-\frac{r}{100}+86=\frac{8600-r}{100}\)

Now, the revenue $R$ obtained is the product of the number of units filled and the monthly rent, thus we may state:

\(\displaystyle R(r)=r\cdot u(r)=\frac{r(8600-r)}{100}\)

To maximize the revenue, given that it is a parabolic function opening downward, we may find the axis of symmetry midway between the roots, at:

\(\displaystyle r=4300\)

And so:

\(\displaystyle R_{\max}=R(4300)=184900\)

Thus, we have found that when the rent per unit is \$4300 per month, the monthly revenue is maximized at \$184,900.

2.) We want to set $h(t)=250$ and then solve for $t$:

\(\displaystyle -5t^2+100t=250\)

Divide through by -5, then add 50 to both sides to get the quadratic in standard form:

\(\displaystyle t^2-20t+50=0\)

Apply the quadratic formula:

\(\displaystyle t=\frac{20\pm\sqrt{(-20)^2-4(1)(50)}}{2(1)}=\frac{20\pm\sqrt{200}}{2}=10\pm5\sqrt{2}=5\left(2\pm\sqrt{2} \right)\)

The smaller root is for when the bullet is on its way up, and the larger root is for when the bullet is on its way back down.

3.) Let's let $v$ be the original slower speed and $t$ be the time it takes at this speed. Since the distance is the same in both cases, then we may write (using $d=vt$):

\(\displaystyle d=vt=(v+9)(t-2)\)

Expanding the right side, we have:

\(\displaystyle vt=vt-2v+9t-18\)

Add $2v-vt$ to both sides:

\(\displaystyle 2v=9t-18\)

Multiply through by $v\ne0$:

\(\displaystyle 2v^2=9vt-18v\)

Since \(\displaystyle d=vt\) we may now arrange this in standard form as:

\(\displaystyle 2v^2+18v-9d=0\)

Applying the quadratic formula, we find:

\(\displaystyle v=\frac{-18\pm\sqrt{18^2-4(2)(-9d)}}{2(2)}=\frac{3\left(-3\pm\sqrt{2d+9} \right)}{2}\)

Since we should assume that $0<v$, we take only the positive root:

\(\displaystyle v=\frac{3\left(\sqrt{2d+9}-3 \right)}{2}\)