Maximize/Minimize Quadratic Functions Help
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.