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Section 3.4 – Build Quadratic Models from Verbal Descriptions
Mathematical models that involve functions are often used to solve real-life problems. The functions must be built
based on the information given in the problem. You must be able to translate the verbal description into a
mathematical model. Assigning symbols to represent the independent and dependent variables and then writing the
function that relates these variables performs this translation.
When a mathematical model is in the form of a quadratic function, the properties of the graph of the function can
provide important information about the model. In particular, you 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 you to answer questions involving optimization – that is, finding the maximum or minimum
values in models.
Build Quadratic Models from Verbal Descriptions
Example 1: The perimeter of a rectangle is 80 feet. Express its area A as a function of the side length L.
w P = 2 w+ 2 L Area =Lw
80 = 2 w+ 2 L =L ( 40 −L)
L 40 =w+L L 40 L
2 =− +
w = 40 −L So, the area A as a function of side length L isA (L) L 40 L
2 =− +
dependent variable independent variable
In economics, revenue R, in dollars, is defined as the amount of money received from the sale of an item and is equal
to the unit selling price p , in dollars, of the item times the number x of units actually sold. That is, R =xp.
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. When the demand equation is linear, the revenue model is a quadratic
function.
Example 2: Suppose unit selling price p and the number x of units sold are related by the demand equation
x 40 , 0 x 300. 5
p (x) +
a) Express the revenue R as a function of the number x of units sold.
Since R =xpand x 40 , 5
p (x) +
= it follows that
R (x)=xp(x)
= x 40 5
x
x 40 x 5
b) Graph the revenue function R(x). What quantity x maximizes the revenue? What is the
maximum revenue?
It takes three points to graph a parabola. Here, use the maximum point and the x-intercepts.
How do you calculate the maximum point? Find the x-intercepts:
2 a
b x (^) v
= R (xv )=R( 100 ) SetR (x)= 0
So, the maximum is at( 100 , 2000 ).
x 40 0 5
x =
x 40 x 0 5
x 40 0 5
x = 0 or
x − 200 = 0
x = 200
So, the x-intercepts are at ( 0 , 0 )and( 200 , 0 ).
50 100 150 200 250 Quantity
Revenue
Section 3.4 – Build Quadratic Models from Verbal Descriptions (continued)
The maximum is at ( 100 , 2000 ). Selling 100 units will give you a maximum revenue of $2000.
Example 3 : The marketing department at Texas Instruments has found that when certain calculators are sold at a
price of p dollars per unit, the number x of calculators sold is given by the demand equation
x(p) = 21 ,000 −150p.
a) Find a model that expresses the revenue R as a function of the price p.
b) What is the domain of R?
c) What unit price should be used to maximize revenue?
d) If this price is charged, what is the maximum revenue?
e) How many units are sold at this price?
f) Graph R.
g) What price should Texas Instruments charge to collect at least $675,000 in revenue?
Solution: a) The Revenue R is R(p) =x(p)p, where x(p) = 21 ,000 −150p.
R(p) =x(p)p
= (21,000 −150p)p
2 = − 150p + 21 ,000p
b) Because x represents the number of calculators sold, you have x 0,so 21 ,000 − 150p 0.
Solving this linear inequality 21 ,000 − 150p 0
21 ,000 150p
p 150
140 p
gives p 140. In addition, Texas Instruments will charge only a positive price for the calculator,
so p 0 too. Combining these inequalities gives the domain of R, which is p| 0 p 140 .
c) The function R is a quadratic function with a = −150,b = 21 ,000,and c =0. Because a 0,the
vertex is the highest point on the parabola. The revenue R is a maximum when the price p is
b p 2a
p 2( 150)
p 300
p =$70.
R(x)
x
Section 3.4 – Build Quadratic Models from Verbal Descriptions (continued)
Example 4 : 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: Figure 25 illustrates the situation. The available fence represents the perimeter of the rectangle.
If x is the length and w is the width, then
2x + 2w =P
2x + 2w = 2000
The area A of the triangle isA =lw
A =xw
To express area A in terms of a single variable, solve the
equation 2x + 2w = 2000 for w and substitute the result in A =xw.
2x + 2w = 2000
2w = −2x + 2000
2w 2x 2000
2 2
w = −x + 1000
So, the area function is A(x) =xw
A(x) = x( −x +1000)
2 A(x) = −x +1000x
Because a 0,the vertex is a maximum point on the graph of A(x). The maximum value occurs at
2 a
b x (^) v
Figure 26 shows the graph of
2 A(x) = − x +1000x.
The maximum value of A(x) isA(x )v =A(500)
2 = − (500) +1000(500)
So, the largest rectangle that can be enclosed by 2000 yards of fencing has an area of 250,000 square
yards. Its dimensions are 500 by 500 yards.
Section 3.4 – Build Quadratic Models from Verbal Descriptions (continued)
Example 5: Analyzing the Motion of a Projectile
A projectile is fired from a cliff 500 feet above the water at an inclination of 45 to the horizontal, with a
muzzle velocity of 400 feet per second. From physics, the height h of the projectile above the water can
be modeled by
2
2
32x h(x) x 500 (400)
= + + , where x is the horizontal distance of the projectile from the base
of the cliff. See Figure 27.
a) Find the maximum height of the projectile.
b) How far from the base of the cliff will
the projectile strike the water?
Solution: a) The height of the projectile is given by a quadratic function, 2
2
32x h(x) x 500 (400)
2 32x x 500 160,
2 x x 500 5000
You are looking for the maximum value of h. Because a 0,the maximum value occurs at the
vertex, whose x-coordinate is 2 a
b x (^) v
The maximum height of the projectile is
2 (2500) h(2500) 2500 500 5000
= 1750 feet
b) The projectile will strike the water when its height is zero. To find the distance x traveled, solve the
equation h(x) = 0.
h(x) = 0
2 x x 500 0 5000
The discriminant is ( )
b 4ac 1 4 (500) 5000
