Introduction to Maximize/Minimize Quadratic Functions
The techniques to maximize/minimize quadratic functions can be applied to problems outside of mathematics. We can maximize the enclosed area, minimize the surface area of a box, maximize revenue, and optimize many other problems. In the first group of problems, the functions to be optimized will be given. In the second, we will have to find the functions based on the information given in the problem. The answers to every problem below will be one or both coordinates of the vertex.
Examples

The weekly profit function for a product is given by P ( x ) = − 0.0001 x ^{2} + 3 x − 12,500, where x is the number of units produced per week, and P ( x ) is the profit (in dollars). What is the maximum weekly profit? How many units should be produced for this profit?
The profit function is a quadratic function which has a maximum value. What information does the vertex give us? h is the number of units needed to maximize the weekly profit, and k is the maximum weekly profit.
Maximize the weekly profit by producing 15,000 units. The maximum weekly profit is $10,000.

The number of units of a product sold depends on the amount of money spent on advertising. If y = −26 x ^{2} + 2600 x + 10,000 gives the number of units sold after x thousands of dollars is spent on advertising, find the amount spent on advertising that results in the most sales.
h will give us the amount to spend on advertising in order to maximize sales, and k will tell us the maximum sales level. We only need to find h .
$50 thousand should be spent on advertising to maximize sales.
Maximum Height
The height of an object propelled upward (neglecting air resistance) is given by the quadratic function s ( t ) = − 16 t ^{2} + v _{0} t + s _{0} , where s is the height in feet, and t is the number of seconds after the initial thrust. The initial velocity (in feet per second) of the object is v _{0} , and s _{0} is the initial height (in feet) of the object. For example, if an object is tossed up at the rate of 10 feet per second, then v _{0} = 10. If an object is propelled upward from a height of 50 feet, then s _{0} = 50. If an object is dropped, its initial velocity is 0, so v _{0} = 0.
Examples

An object is tossed upward with an initial velocity of 15 feet per second from a height of four feet. What is the object’s maximum height? How long does it take the object to reach its maximum height?
Because the initial velocity is 15 feet per second, v _{0} = 15, and the initial height is four feet, so s _{0} = 4. The function that gives the height of the object (in feet) after t seconds is s ( t ) = − 16 t ^{2} + 15 t + 4.
The object reaches its maximum height of 7.515625 feet after 0.46875 seconds.

A projectile is fired from the ground with an initial velocity of 120 miles per hour. What is the projectile’s maximum height? How long does it take to reach its maximum height?
Because the projectile is being fired from the ground, its initial height is 0, s _{0} = 0. The initial velocity is given as 120 miles per hour—we need to convert this to feet per second. There are 5280 feet per mile, so 120 miles is 120(5280) = 633,600 feet. There are 60(60) = 3600 seconds per hour.
Now we have the function: s ( t ) = − 16 t ^{2} + 176 t + 0 = − 16 t ^{2} + 176 t.
The projectile reaches its maximum height of 484 feet after 5.5 seconds.
Finding the Maximum Height and Horizontal Distance
Another problem involving the maximum vertical height is one where we know the horizontal distance traveled instead of the time it has traveled. The x coordinates describe the object’s horizontal distance, and the y coordinates describe its height. Here we will find the maximum height and how far it traveled horizontally to reach the maximum height.
Example

A ball is thrown across a field. Its path can be described by the equation y = − 0.002 x ^{2} + 0.2 x + 5, where x is the horizontal distance (in feet) and y is the height (in feet). See Figure 6.8. What is the ball’s maximum height? How far had it traveled horizontally to reach its maximum height?
Fig. 6.8
k will answer the first question, and h will answer the second.
The ball reached a maximum height of 10 feet when it traveled 50 feet horizontally.
Maximum Revenue
The revenue of a product or service can depend on its price in two ways. An increase in the price means that more revenue per unit is earned but fewer units are sold. A decrease in the price means that less revenue is earned per unit but more units are sold. Quadratic functions model some of these relationships. In the next problems, a current price and sales level are given. We will be told how a price increase or decrease affects the sales level. We will let x represent the number of price increases/decreases. Suppose every $10 decrease in the price results in an increase of five customers. Then the revenue function is (old price − 10 x )(old sales level + 5 x ). If every $50 increase in the price results in a loss of one customer, then the revenue function is (old price + 50 x )(old sales level − 1 x ). These functions are quadratic functions which have a maximum value. The vertex tells us the maximum revenue and how many times to decrease/increase the price to get the maximum revenue.
Examples

A management firm has determined that 60 apartments in a complex can be rented if the monthly rent is $900, and that for each $50 increase in the rent, three tenants are lost with little chance of being replaced. What rent should be charged to maximize revenue? What is the maximum revenue?
Let x represent the number of $50 increases in the rent. This means if the rent is raised $50, x = 1, if the rent is increased $100, x = 2, and if the rent is increased $150, x = 3. The rent function is 900 + 50 x . The number of tenants depends on the number of $50 increases in the rent. So, if the rent is raised $50, there will be 60 − 3(1) tenants; if the rent is raised $100, the there will be 60 − 3(2) tenants; and if the rent is raised $150, there will be 60 − 3(3) tenants. If the rent is raised $50x, there will be 60 − 3 x tenants. The revenue function is
The maximum revenue is $54,150. Maximize revenue by charging 900 + 50(1) = $950 per month for rent.

A cinema multiplex averages 2500 tickets sold on a Saturday when ticket prices are $8. Concession revenue averages $1.50 per ticket sold. A research firm has determined that for each $0.50 increase in the ticket price, 100 fewer tickets will be sold. What is the maximum revenue (including concession revenue) and what ticket price maximizes the revenue?
Let x represent the number of $0.50 increases in the price. The ticket price is 8 + 0.50 x . The average number of tickets sold is 2500 − 100 x . The average ticket revenue is (8.00 + 0.50 x )(2500 − 100 x ). The average concession revenue is 1.50(2500 − 100 x ). The total revenue is
To maximize revenue, the ticket price should be 8.00 + 0.50(3) = $9.50, and the maximum revenue is $24,200.

The manager of a performing arts company offers a group discount price of $45 per person for groups of 20 or more and will drop the price by $1.50 per person for each additional person. What is the maximum revenue? What size group will maximize the revenue?
Because the price does not change until more than 20 people are in the group, we will let x represent the additional people in the group. What is the price per person if the group size is more than 20? If one extra person is in the group, the price is 45 − 1(1.50). If there are two extra people, the price is 45 − 2(1.50); and if there are three extra people, the price is 45 − 3(1.50). So, if there are x additional people, the price is 45 − 1.50 x . The revenue is
The group size that maximizes revenue is 20 + 5 = 25. The maximum revenue is $937.50.
Maximum Enclosed Area
Optimizing geometric figures are common calculus and precalculus problems. In many of these problems, there are more than two variables. We will be given enough information in the problem to eliminate one of the variables. For example, if we want the area of a rectangle, the formula is A = LW . If we know the perimeter is 20, then we can use the equation 2 L + 2 W = 20 to solve for either L or W and substitute this quantity in the area function, reducing the equation from three to two variables. The new area function will be quadratic.
Examples

A parks department has 1200 meters of fencing available to enclose two adjacent playing fields. (See Figure 6.9.) What dimensions will maximize the enclosed area? What is the maximum enclosed area?
Fig. 6.9
The total enclosed area is A = LW . Because there is 1200 meters of fencing available, we must have L + W + W + W + L = 1200 (see Figure 6.10 ).
Fig. 6.10
We can solve for L or W in this equation and substitute it in A = LW , reducing the equation to two variables. We will solve for L in 2 L + 3 W = 1200.
Now A = LW becomes . This function has a maximum value.
The width that maximizes the enclosed area is 200 meters, the length is meters. The maximum enclosed area is 60,000 square meters.
Another common fencing problem is one where only three sides of a rectangular area needs to be fenced. The fourth side is some other boundary like a stream or the side of a building. We will call two sides W and the third side L . Then “2 W + L = amount of fencing” allows us to solve for L and substitute “ L = amount of fencing −2 W ” in A = LW to reduce the area formula to two variables.
Example

A farmer has 1000 feet of fencing materials available to fence a rectangular pasture next to a river. If the side along the river does not need to be fenced, what dimensions will maximize the enclosed area? What is the maximum enclosed area?
Fig. 6.11
Using the fact that 2 W + L = 1000, we can solve for L and substitute this quantity in the area formula A = LW .
2 W + L = 1000
L = 1000 − 2 W
A = LW
A = (1000 − 2 W ) W = −2 W ^{2} + 1000 W
This quadratic function has a maximum value.
Maximize the enclosed area by letting W = 250 feet and L = 1000 − 2(250) = 500 feet. The maximum enclosed area is 125,000 square feet.
In the last problems, we will maximize the area of a figure but will have to work a little harder to find the area function to maximize.
Examples

A window is to be constructed in the shape of a rectangle surmounted by a semicircle (see Figure 6.12). The perimeter of the window needs to be 18 feet. What dimensions will admit the greatest amount of light?
Fig. 6.12
The dimensions that will admit the greatest amount of light are the same that will maximize the area of the window. The area of the window is the rectangular area plus the area of the semicircle. The area of the rectangular region is LW . Because the width of the window is the diameter (or twice the radius) of the semicircle, we can rewrite the area as L (2 r ) = 2 rL. The area of the semicircle is half of the area of a circle with radius r, . The total area of the window is
Now we will use the fact that the perimeter is 18 feet to help us replace L with an expression using r . The perimeter is made up of the two sides (2L) and the bottom of the rectangle (2 r ) and the length around the semicircle. The length around the outside of the semicircle is half of the circumference of a circle with radius r , . The total perimeter is P = 2 L + 2 r + π r . This is equal to 18. We will solve the equation 2 L + 2 r + π r = 18 for L.
Now we will substitute for L in the area formula.
This quadratic function has a maximum value.
Maximize the amount of light admitted in the window by letting the radius of the semicircle be about 2.52 feet, and the length about 2.52 feet.

A track is to be constructed so that it is shaped like Figure 6.13, a rectangle with a semicircle at each end. If the inside perimeter of the track is to be mile, what is the maximum area of the rectangle?
Fig. 6.13
The length of the rectangle is L . Its width is the diameter of the semicircles (or twice their radius). The area formula for the rectangle is A = LW = L (2 r ) = 2 rL . The perimeter of the figure is the two sides of the rectangle (2 L ) plus the length around each semicircle (π r ). The total perimeter is 2 L + 2π r . Although we could work with the dimensions in miles, it will be easier to convert 1/4 mile to feet. There are 5280/4 = 1320 feet in 1/4 mile.
We will solve 2 L + 2π r = 1320 for L . Solving for r works, too.
The area function has a maximum value.
The maximum area of the rectangular region is about 69,328 square feet.

A rectangle is to be constructed so that it is bounded below by the x axis, on the left by the y axis, and above by the line y = −2 x + 12. (See Figure 6.14). What is the maximum area of the rectangle?
Fig. 6.14
The coordinates of the corners will help us to see how we can find the length and width of the rectangle.
Fig. 6.15
The height of the rectangle is y and the width is x . This makes the area A = xy . We need to eliminate x or y . Because y = −2 x + 12, we can substitute − 2 x + 12 for y in A = xy to make it the quadratic function A = xy = x (−2 x + 12) = −2 x ^{2} + 12 x .
The maximum area is 18 square units.
Maximize/Minimize Quadratic Functions Practice Problems
Practice
 The average cost of a product can be approximated by the function C ( x ) = 0.00025 x ^{2} − 0.25 x + 70.5, where x is the number of units produced and C ( x ) is the average cost in dollars. What level of production will minimize the average cost?
 A frog jumps from a rock to the shore of a pond. Its path is given by the equation , where x is the horizontal distance in inches, and y is the height in inches. What is the frog’s maximum height? How far had it traveled horizontally when it reached its maximum height?
 A projectile is fired upward from a tenfoot platform. The projectile’s initial velocity is 108 miles per hour. What is the projectile’s maximum height? When will it reach its maximum height?
 Attendance at home games for a college basketball team averages 1000 and the ticket price is $12. Concession sales average $2 per person. A student survey reveals that for every $0.25 decrease in the ticket price, 25 more students will attend the home games. What ticket price will maximize revenue? What is the maximum revenue?

A school has 1600 feet of fencing available to enclose three playing fields (see Figure 6.16). What dimensions will maximize the enclosed area?
Fig. 6.16

The manager of a large warehouse wants to enclose an area behind the building. He has 900 feet of fencing available. What dimensions will maximize the enclosed area? What is the maximum area?
Fig. 6.17

A swimming pool is to be constructed in the shape of a rectangle with a semicircle at one end (see Figure 6.12). If the perimeter is to be 120 feet, what dimensions will maximize the area? What is the maximum area?

A rectangle is to be constructed so that it is bounded by the x axis, the y axis, and the line y = −3 x + 4 (see Figure 6.18). What is the maximum area of the rectangle?
Fig. 6.18
Solutions
 We only need to find h .
Minimize the average cost by producing 500 units.

k answers the first question, h answers the second.
The frog reached a maximum height of 10 inches and had traveled 12 inches horizontally when it reached its maximum height.

The formula s ( t ) = − 16 t ^{2} + v _{0} t + s _{0} is in feet and seconds, so we need to convert 108 miles per hour to feet per second. There are 5280 feet in a mile and 60(60) = 3600 seconds in an hour.
Replacing v _{0} with 158.4 and s _{0} with 10, we have the function giving the height of the projectile after t seconds, s ( t ) = − 16 t ^{2} + 158.4 t + 10.
The projectile reaches a maximum height of 402.04 feet after 4.95 seconds.

We will let x represent the number of $0.25 decreases in the ticket price. The ticket price is 12 − 0.25 x and the average number attending the games is 1000 + 25 x . Ticket revenue is (12 − 0.25 x )(1000 + 25 x ). Revenue from concession sales is 2(1000 + 25 x ). Total revenue is
The ticket price that will maximize revenue is 12 − 0.25(8) = $10 and the maximum revenue is $14,400.

The total area is A = L W . Because there is 1600 feet of fencing available, 2 L + 4 W = 1600. Solving this equation for L , we have L = 800 − 2 W . Substitute 800 − 2 W for L in A = LW .
Maximize the enclosed area by letting the width be 200 feet and the length be 800 − 2(200) = 400 feet.

The enclosed area is A = LW . Because 900 feet of fencing is available, 2 W + L = 900. Solving this for L , we have L = 900 − 2 W . We will substitute 900 − 2 W for L in A = LW .
A = LW
= (900 − 2 W ) W = −2 W ^{2} + 900 W
Maximize the enclosed area by letting the width be 225 feet and the length 900 − 2(225) = 450 feet. The maximum enclosed area is 101,250 square feet.

The area of the rectangle is 2 rL (the width is twice the radius of the semicircle). The area of the semicircle is half the area of a circle with radius r , . The total area of the pool is . After finding an equation for the perimeter, we will solve the equation for L and substitute this for L in . The perimeter of the rectangular part is L + 2 r + L = 2 r + 2 L . The length around the semicircle is half the circumference of a circle with radius r , . The total length around the pool is 2 L + 2 r + π r which equals 120 feet.
Maximize the area by letting the radius of the semicircle be about 16.8 feet and the length of the rectangle about feet. The maximum area is about 1008.2 square feet.

The area is A = LW . The length of the rectangle is y (the distance from (0,0) and (0, y )). The width is x (the distance from (0, 0) and ( x , 0)). The area is now A = xy . Because y = −3 x + 4, we can substitute − 3 x + 4 for y in A = xy .
The maximum area is square units.
Practice problems for this concept can be found at: Quadratic Functions Practice Test.