Week 9 – Precalculus 11

This week in precalc 11 we learned about how to find maximum values and revenue using quadratic functions, we can also find different variables depending on the word problem for example if there’s a question about area then the question may also ask for the length and width and we have to use the quadratic function to solve for the side lengths along with the maximum values.

Example: A company can sell 56 phones, each one being sold for $300. For every $30 the price increases, the number of phones sold lower by 3. What is the maximum revenue that the company can make by selling the phones?

The formula we use for revenue is: (price)(# of sales)

so knowing this lets see what we know

  • we have 60 phones
  • each one is sold for $300
  • the price can increase by $30
  • But then the number of phones drops by 3
  • Were trying to find the maximum value

know we need to use a variable to find out our maximum revenue

The variables will be how many phones we decrease by and how much we increase the price

So lets plug in the number we know into the formula

(300 + 30x)(60 – 3x)

Now we can solve for the variables by completing the square but first we distribute

(300 + 30x)(60 – 3x)

18,000 – 900x + 1,800x – 90x^2

-90x^2 + 600x + 15,000

Now we can complete the square and find our vertex which will give us our “x” value and maximum revenue.

\frac{-90x^2}{-90} +\frac{900x}{-90} + 18,000

-90(x^2 – 10 + 25 – 25) + 18,000

-90(x - 5)^2 + 18,000

Vertex is : (5, 18,000)

Our “5” shows our value for “x” and the 18,000 shows our maximum value but because in this question it’s asking for revenue it is our maximum revenue.

Sometimes a question may ask, “which price for a phone will maximise the revenue?”

and what you do is go back to our formula with the numbers

(300 + 30x)(60 – 3x)

and in this case the “(300 + 30x)” talks about the price for a single phone so we plug in our “x” value we got from our vertex and solve to find our the price of one phone that would maximise the revenue.

(300 + 30x)

(300 + 30(5))

(300 + 150) = 450

so this means the price of $450 for one phone will maximise the revenue in this question.

 

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