# Thread: Trying to solve Alg 2 problem

1. ## Trying to solve Alg 2 problem

I'm trying to help a student with a word problem and am stuck on it. He's learning how to use functions to find maximum and minimum values for word problems. None of the examples in his text are like the problem below, and we can't figure it out.:

"A publishing company can get 500 subscribers for a weekly news magazine if the yearly subscription rate is $60. The company will get 10 more subscribers for each$1 decrease in the yearly rate. What rate will produce the maximum income, and what will that income be?"

In other problems of this type, we'd say the following:
500 + x = total subscribers
60-x = subscription rate
Then if you solve for the f(x) = (500+x)(60-x) you get an equation that tells the maximum income and also the maximum income. But, in this case it's not $1 per additional subscriber, but per 10 subscribers. What am I doing wrong in my equation then? Thanks!! 2. Originally Posted by hereami I'm trying to help a student with a word problem and am stuck on it. He's learning how to use functions to find maximum and minimum values for word problems. None of the examples in his text are like the problem below, and we can't figure it out.: "A publishing company can get 500 subscribers for a weekly news magazine if the yearly subscription rate is$60. The company will get 10 more subscribers for each $1 decrease in the yearly rate. What rate will produce the maximum income, and what will that income be?" In other problems of this type, we'd say the following: x = additional subscribers 500 + x = total subscribers 60-x = subscription rate Then if you solve for the f(x) = (500+x)(60-x) you get an equation that tells the maximum income and also the maximum income. But, in this case it's not$1 per additional subscriber, but per 10 subscribers. What am I doing wrong in my equation then?
Thanks!!
let x = number of $1 decreases in the subscription rate 500+10x = number of subscribers 60-x = subscription rate income ... f(x) = (500+10x)(60-x) max income at x = 5 max income ... f(5) = (550)(55) =$30250

3. ## still unsure...

Okay, so I understand how you set it up. But, how did you solve f(x) = (500+10x)(60-x)?

I get f(x) = 30,000 - 500x +600x -10x 2
=-10x 2 +100x + 30,000
= x 2-10x - 3,000

Right so far? Then if I complete the square I get
=(x-5)2 -3025.
But that's not what you ended up with. What am I missing?

4. the function is quadratic, and its graph is a parabola that opens down.

setting f(x) = 0 ...

(500+10x)(60-x) = 0

x-intercepts at x = -50 and x = 60

because of the parabola's symmetry, the vertex is located midway between the two x-intercepts ... at x = 5.

that's where f(x) has its maximum.