# Quadratic Equations : Maximum and Minimum Values

From this we can see that we have function that whose graph is a parabola and we want the values of w that make the function positive. The parabola opens downward, the vertex is at Let me show you an actual example of inequalities. Several years ago I was the chairman of the banquet committee for our ski club. Since one of our members was a member of the local yacht club, we decide to hold our annual banquet and dance at the yacht club. The area we were renting was part of the main room. The yacht club wanted to use part of the room for their members activities and we wanted a private area. We were located in one corner of the room where there were two glass walls overlooking the harbor. The club had 105 linear feet of temporary walls that we could set up to form a rectangular room. Fire regulations required that we have at least 20 square feet of space for each attendee. We expected 100 to 130 people to attend the function. I had to determine if we had enough temporary walls to accommodate our requirements or if I would have to rent more. Here is where inequalities come into play.

First, I let W equal the length of one wall.

Then the area of the room I can form is:

I had to plan on the maximum number of people so I need at least 130x20=2600 sq ft. Now I can write my inequality:

and the x-axis intercepts are at 40 and 65.

The parabola is:

This means that I do have enough temporary walls. If I make the room 65 ft x 40 ft the area is just 2600 square feet and I have used the entire temporary wall. Can I make the room larger if I use all of the wall sections and change the dimensions? The answer is yes, because according to the graph I can select any value between 40 and 65 for w and be above the fire code limit of 2600 sq ft.

This is a two-part question. Without solving this problem, what would you expect the value of W to be to maximize the banquet area?

If I had to pay for the amount of wall I was using, how would the problem be set up differently?

#### Solution Preview

Hi there

If the length of one wall is x then teh length of teh other wall section is (105-x) and the area is:

S=x*(105-x)=-x^2+105x

And this area must be larger than 2600 sq. ft. So teh inequality has the form of a parabola:

-x^2+105x>2600

-x^2+105x-2600>0

A parabola is symmetric about its maximal point (or minimal if the parabola opens up).

Therefore the parabola that represents the total area of the room as a function of one side (and it is equal ...

#### Solution Summary

A parabola is analyzed and applied to a maximum value problem.