Maximize/Minimize Quadratic Functions Help (page 2)
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.
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.
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.
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.
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?
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.
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.
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.
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?
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 ).
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.
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