# Quadratic Models in Precalculus

These are my notes on quadratic models in precalculus.

When a mathematical model leads to a quadratic function, the properties of the

graph of the quadratic function can provide important information about the

model. We can use the quadratic function to determine the maximum or minimum

value of the function. The fact that the graph of a quadratic function has a

maximum or minimum value enables us to answer questions involving optimization

or finding the maximum or minimum values involving quadratic functions.**Solving Applied Problems**

In economics, revenue is defined as the amount of money received from the sale

of an item and is equal to the unit selling price of the item times the number

of units that were sold.

\(R=xp\)

In economics, the Law of Demand states that p and x are related. As one

increases, the other decreases. The equation that relates p and x is called the

demand equation.**Example 1**

The marketing department at Texas Instruments has found that, when certain

calculators are sold at a certain price, the number of calculators sold is given

by the demand equation:

\(x=21000-150p\)

1. Express the revenue as a function of price

2. What unit price should be used to maximize revenue

3. If this price is charged, what is the maximum revenue

4. How many units are sold at this price

5. Graph

Solution:

1. The revenue is \(R=xp\) where \(x=21000-150p\).

\((21000-150p)p= -150p^2+21000p\)

2. The function is a quadratic function with a=-150, b=21000, and c=0.

Because a<0, the vertex is the highest point on the parabola.

The revenue is therefore a maximum when the price is:

\(p=\frac{-b}{2a} = -\frac{21000}{2(-159)} = \frac{21000}{-300} = $70.00\)

3. The maximum revenue is:

\(R(70) = -150(70)^2 + 21000(70) = $735000\)

4. The number of calculators sold is given by the demand equation:

\(x=21000-150p\)

At a price of \(p=$70\),

\(x=21000-150(70) = 10500\)**Example 2**

Maximizing the Area Enclosed by a Fence

A farmer has 2000 yards of fence to enclose a rectangular field. What are the

dimensions of the rectangle that encloses the most area?

Solution:

The available fence represents the perimeter of the rectangle. If x is the

length and w is the width, then:

\(2x+2w=2000\)

The area of the rectangle is:

\(a=xw\)

To express area in terms of a single variable, we solve the equation for w and

substitute the result in \(a=xw\). Then area involves only the variable x. You

could also solve the equation for x and express area in terms of w alone.

\(2x + 2w = 2000\)

\(2w = 2000-2x\)

\(w = \frac{2000-2x}{2} = (1000-x)\)

So, the area is:

\(a = xw = x(1000-x) = -x^2 + 1000x\)

Now, area is a quadratic function of x.

\(a(x)=-x^2+1000x\)

When you graph this, you see a<0, so the vertex is a maximum point on the graph.

The maximum value occurs at:

\(x=-\frac{b}{2a} = -\frac{1000}{2(-1)} = 500\)

The maximum value of a is:

\(a(-\frac{b}{2a}) = a(500) = -500^2 + 1000(500) = -250000 + 500000 = 250000\)

The largest rectangle that can be enclosed by 2000 yards of fence has an area of

250000 square yards. Its dimensions are 500 by 500 yards.**Example 3**

Analyzing the motion of a projectile

A projectile is fired from a cliff 500 feet above the water at an inclination of

45 degrees to the horizontal, with a muzzle velocity of 400 feet per second. In

physics, it is established that the height of the projectile above the water is

given by:

\(h(x)=\frac{-32x^2}{(400)^2} + x + 500\)

X is the horizontal distance of the projectile from the base of the cliff.

1. Find the maximum height of the projectile

2. How far from the base of the cliff will the projectile strike the water?

Solution:

1. The height of the projectile is given by a quadratic function.

\(h(x)=\frac{-32x^2}{(400)^2}+x+500=\frac{-1}{5000}x^2+x+500\)

We are looking for the maximum value of h. Since a<0, the maximum value is

obtained at the vertex.

\(x=-\frac{b}{2a}=-\frac{1}{2(-1/5000)}=\frac{5000}{2}=2500\)

The maximum height of the projectile is:

\(h(2500)=\frac{-1}{5000}(2500)^2+2500+500=-1250+2500+500=1750ft\)

2. The projectile will strike the water when the height is zero. To find the

distance x traveled, we need to solve the equation:

\(h(x)=\frac{-1}{5000}x^2+x+500=0\)

We find the descriminant first.

\(b^2-4ac=1^2-4(\frac{-1}{5000})(500)=1.4\)

Then: \(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-1 \pm

\sqrt{1.4}}{2(-1/5000)}=5458ft\)