Max/Min Problems

One of the many real-life applications of quadratic functions is performing maximum/minimum studies in certain situations. On this page we will work through an example, give the steps for solving a max/min problem, and then look at one more example. Remember, these examples are not meant to replace your notes or what your teacher did in class.

Example 1

If a car dealership sets the price of their cars at $28 000 they will sell 54 cars. Every time they drop the price $1000, 2 more cars will be sold. What should the price of the cars be set at to maximize sales?

Solution

This question is concerned with maximizing sales. How are total sales calculated? It is calculated by multiplying the price of the item being sold by the number that is sold. In the example above, the total sales before changes are:

Total Sales = Price(Number Sold)
Total Sales = 28000(54)
Total Sales = 1512000

Is this the maximum amount of sales that can be generated? We don't know yet. We have to do more work. In the question you are told what happens when the price is dropped by $1000. Let's call x the number of $1000 drops in price. This means that the new cost per car will be $28000 - $1000x ($28000 minus $1000 for each 'drop in price'). If we drop the price by $1000, 2 more cars will be sold. That means the new number of cars sold is 54+2x (54 plus 2 for every $1000 drop in price (x)). This makes our new sales calculation:

Total Sales = 'New Price'('New Number Sold')
Total Sales = (28000 - 1000x)(54+2x)

What can we do with this equation? Well, you can multiply it out to get:

Total Sales = 28000 + 56000x - 54000x - 2000x²
rearrange and simplify:
Total Sales = -2000x² +2000x + 28000

YIKES! That is one HUGE quadratic! This is what will tell us what we need to do to the price in order to maximize sales. You can see that since the a value of this quadratic is negative (-2000) that this will be a downward-turning parabola. A quick sketch of this parabola is shown below, with the maximum point identified.

The quick algebraic way to find the x-coordinate of the vertex is to use the expression . For this quadratic we get x = -2000/-4000 (or x = .5) This is the x-coordinate of the maximum point on the graph. We are looking for maximum sales. This value for x (.5) is the number of $1000 drops in price we need to maximize the sales. That means we need half of a $1000 drop in price, so we should drop the price by $500.

Therefore, the price that the cars should be set at to maximize sales is $27500.

General Steps for Solving a Max/Mix Problem

1. Start by determining WHAT needs to be maximized or minimized. Sales? Profit? Yield? Area? Make sure you know how to calculate the thing that needs to be maximized. (For example, if you need to maximize an area, make sure you know that A = bh).
2. Set up an equation that calculates what you are looking for before making any changes.
3. Change both parts of the formula to reflect what will happen when you make changes. This includes defining what 'x' will stand for (in the last example it was the number of $1000 drops in price).
4. Multiply your new formula (which has x's in it) to arrive at a quadratic function.
5. Find the x-coordinate of the vertex (which is located at the maximum or the minimum of the function) using .
6. Use the x-value you just found to answer the question.

Example 2

A rectangular field beside a river is to be fenced. No fence is needed along the river bank. What are the dimensions of the field of maximum area which can be enclosed using 80 m of fencing?

Solution

It always helps to draw a diagram of the situation, if you can. A picture of this situation is shown below:

Let's follow the steps that are outlined above.

1. The quantity that we want to maximize is the area. The formula for area is length times width.
2. The proper formula is A = lw.
3. Now we have to set up the above equation for this situation. What makes this question different is the fact that we are looking at two variables above. We have to set up an equation in one variable. So we will still call the width w, but we will try to define the length (l) in terms of the width (w).
   
   Since we have 80 m of fencing to work with, and we have already used 2w of this fencing to make the widths, we have (80 - 2w) m of fencing left to fence the width. So now our formula looks like:
   
   A = (80 - 2w)w or,
4. A = -2w² + 80w.
5. The x-coordinate (or in this case, the w-coordinate) of the vertex is:
   w = -b/2a
   w = -80/2(-2)
   w = -80/(-4)
   w = 20
6. Now we know that to acheive a maximum area we have to set the width at 20 and length at 40 (from 80 - 2(20)).

Now that you have seen these examples you can head back to the Quadratic Functions main page.