Problem

In Exercises 71-74, find two positive real number…

03:44

Problem 72 Medium Difficulty

In Exercises 71-74, find two positive real numbers whose product is a maximum.

The sum is $ S $.

Answer

$\frac{S}{2} \times \frac{S}{2}$

Related Courses

Algebra

Precalculus with Limits

Chapter 2

Polynomial and Rational Functions

Section 1

Quadratic Functions and Models

Discussion

You must be signed in to discuss.

Top Algebra Educators

Watch More Solved Questions in Chapter 2

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Problem 93

Problem 94

Problem 95

Problem 96

Problem 97

Problem 98

Problem 99

Problem 100

Video Transcript

let's go ahead. And do you know are two positive real numbers by X and y. Now we're given that the sum of these two is equal to the quantity s and the thing that we want to maximizes the product. So p the fine is the product and this is what we like to maximize right here. So in order to rewrite, this is a function of just one variable. Let's go ahead and replace why by solving this equation up here. So go ahead and subtract X from each side, giving you why Equals s minus X. There we go. And then we can just go ahead and multiply this all out. And there's our quadratic in terms of only one variable. So by looking at this quadratic, you see the negative sign in the front that tells you that if you wanna graft this thing more or less, you get something like this and we're looking for the maximum which will be the y value. So if you're quadratic is of the form X squared plus B x plus C than the Vertex is the point negative b over two a. And then plug that into your function to find the y value. So in our case, we're front where we wanted to know what X value puts us at the Vertex here. So in this case, we should take negative B over two A. So in our case, we have a CZ minus one B is s and then we don't need see here So let's not even bother. So now we'LL have X equals negative B over two A. So that will end up just being s over too. So this is the X value that makes the product the maximum they're not They're not asking for the maximum value that but that would be this value here. So not the difference between finding the eggs and finding the mass value. We're looking for two real numbers. We found one of them X and I will go ahead and find the other. Why? By using this equation here, so plugging in our new X value, we get s minus s over to, which is also it's over too. So are two numbers. There have some as and that have maximum product rs over to in s over too

We have video lessons for 96.97% of the questions in this textbook