
Intermediate Algebra
A tour organizer serves 450 customers in a day, if they charge 10 dollars per person. The company estimates that it will lose 10 customers for every increase of 0.50 dollars in the fare. Find the fare that would yield the maximum income to the company.

Re: Intermediate Algebra
Can you get an equation for the income?

Re: Intermediate Algebra
let the fare be increased by x times half dollar.....
therefore equation becomes
$\displaystyle
f\left( x \right) = (450  10x)(10 + x/2) = 4500  10x + 225x  5{x^2} = 4500 + 215x  5{x^2}\\
\frac{d}{{dx}}(f\left( x \right)) =  10x + 215\\
\frac{d}{{dx}}(f\left( x \right)) = 0\\
 10x + 215 = 0\\
x = 21.5\\
f(21.5) = 6811.25\[\max \]
$
therefore fare which yields max profit is 10+21.5/2=10+10.75=20.75$

Re: Intermediate Algebra
Just neatening things up a bit:
A bit of LaTeX coding advice: Unless you are writing a matrix or array avoid using the returns in the tex lines. That will get rid of those pesky < br/ >s. I also got rid of the \\, but I'm not sure it did anything.
$\displaystyle f\left( x \right) = (450  10x)(10 + x/2) = 4500  10x + 225x  5{x^2} = 4500 + 215x  5{x^2}$
$\displaystyle \frac{d}{{dx}}(f\left( x \right)) =  10x + 215$
$\displaystyle \frac{d}{{dx}}(f\left( x \right)) = 0$
$\displaystyle  10x + 215 = 0$
$\displaystyle x = 21.5$
$\displaystyle f(21.5) = 6811.25\[\max \] $
therefore fare which yields max profit is 10+21.5/2=10+10.75=20.75$
Dan