Lesson 11. Ma February 8 th, 2017

Transcription

1 Lesson 11 Ma February 8 th, 2017 This lesson focuses on applications of quadratics.the nice thing about quadratic expressions is that it is very easy to find their maximum or minimum values, namely their vertex. This will be very useful when try to maximizing a quantity (say area, volume, revenue, etc.) or minimize a quantity (say cost, time, etc.) if it has a quadratic representation. Example 1: The sum of two integers is 30. Find their values in order to maximize their product. Solution: x ` y 30 and we re interested in xy. We use the usual trick to put one variable in terms of the other, Thus we observe that: x ` y 30 ùñ y 30 x xy xp30 xq x 2 ` 30x Some kind of quadratic form that we ve seen before. Thus if we wish to maximize the product, we are really interested in the vertex of the resulting parabola (as it opens downward, the vertex will be a maximum). 15 and conse- Using the h b 30 method we see that the best possible x-value is 2a quently y Thus 2p 1q x 15 and y 15 Note: in a situation like this, both coordinates of the vertex ph, kq will be significant. h will correspond to some input, rate, or value that will optimize the result, while k will be that optimum result. 1

2 Example 2: Say we are given 600 ft. of fencing to enclose a rectangular space divide in two as follows: w Find the dimensions that will maximize the area and find the maximal area. Solution: Again, we do the same trick where we express one variable interms of the other. Because we are given 600 ft. of fencing we have 3w ` 2l 600 ùñ 2l 600 3w ùñ l w Thus we see that the area can be expressed as ˆ A lw w w 3 2 w2 ` 300w Again we have a quadratic and using our vertex equation we see that the width that will give the largest area is w And from the equation for the length (in terms p2q 3 2 of w) that we derived we have l p100q 150. Thus the dimensions that provided 2 the greatest area are w 100 and l 150 Furthermore, the maxiaml area is just the product of the two A , 000 l Note: we may generalize the ideas in this problem to construct a fence with any number of sides (even when sides are removed). 2

3 Example 3: A rental car company has 500 cars on their lot and they can rent them all at the rate $40/day. They also know that for every $1 increase from that rate, they will rent out 5 less cars. Find the rate that will maximize the profit. Solution: In a problem like this, the set-up will always be the most difficult part. Let s see if we can simplify it. In general: Profit prateqp#items sold/rentedq Let x denote the number of $1 increases in the rate from $40 (making the rate 40 ` 1x 40 ` x). Then we can express the number of cars rented by Then the profit is given by Number rented 500 5x Profit p40 ` xqp500 5xq 5x 2 ` 300x ` 20, 000 Now we solve for the x-value (the increase in the rate) that will give the optimal profit. We do the b 2a method p 5q 30 So 30 is the number of $1 increases to the rate which will give the maximum profit. Then 30 $1 $30 is the increase in the rate from $40 so we need to add them to find the actual rate. $40 ` $30 $70 is the rate at which the company will make the most money. 3

4 Example 4: Say a toy company can produce a model train at a production cost of $5. When sold at $28, 400 model trains are sold per a week. For every $0.50 decrease from this rate 20 more are sold. Find the rate that will maximize profit. Solution: In general Profit Sales Profit Total Cost pindividual rateqp#soldq pindividual costqp#soldq pindividual rate individual costqp#soldq Similar to the last problem, let x denote the number of $0.50 decreases. Then we see Number sold 400 ` xp20q 20x ` 400 We know the cost of production is fixed at $5 so now we only need to find an expression for the rate at which they are sold. Since x represents the number of $0.50 decreases, the rate at which they are sold is given by x. Putting this all together we have, Profit pp28 0.5xq 5qp20x ` 400q p23 0.5xqp20x ` 400q 10x 2 ` 260x ` 9200 Now we look for the h-coordinate of the vertex to find the number of $0.50 decreases in the price that will give the maximum profit p 10q 13 Recall that this represents the number of $0.50 decreases in price. Thus the total decrease in price is 13 $0.50 $6.50. Now to find the best price we subtract $28 $6.50 $

5 Example 5: The height of a ball thrown from the top of a building can be modeled by the equation hptq 16t 2 ` 81t ` 300 where h is the height in feet and t is time in seconds. (a) Find the height of the building (b) Find the time it takes for the ball to reach it maximal height (c) Find the maximal height of the ball (d) Find the time the ball hits the ground Solution: (a) The height of the building is the same as the height of the ball at time t 0. hp0q 300 (b) We need to find the h-value of the vertex of the parabola. 81 2p 16q «2.53 (c) The actually height is given by the k-value of the vertex. We only need to evaluate the expression at 2.53 seconds 16p2.53q 2 ` 81p2.53q ` 300 « (d) In orde to find when the ball hits the ground, we need to find when the height of the ball is 0. For what time t is 16t 2 ` 81t ` 300 0? We use the quadratic formula t 81 a81 2 4p 16qp300q 2p 16q We will get two answers but we ignore the negative result as t ě 0 since it represents time. We find t «7.55 5

6 Example 6: The velocity of a subatomic particle moving through space can be modeled by vptq 0.8t 2 3.6t ` 0.3. (a) Find the times the particle is not moving (b) Find when the particle is moving forwards (c) Find when the particle is moving backwards Solution: (a) The particle will not be moving when it has zero velocity so we must solve when 0.8t 2 3.6t ` We use the quadratic formula t 3.6 ap 3.6q 2 4p0.8qp0.3q 2p0.8q t 0.08 and t 4.42 (b) We now want to consider when the velocity is positive or equivalently when the graph is above the x-axis. Since 0.8 ą 0 we know this parabola will open upwards. Since we know what the roots are we see the velocity is negative between the roots and is positive outside of the roots. Since t ě 0 (no such thing as negative time) we have that the particle is moving forwards on r0, 0.08q Y p4.42, 8q (c) Similarly the velocity is negative between the zeroes so the particle is moving backwards on p0.08, 4.42q 6