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Precalculus
Course: Precalculus > Unit 4
Lesson 5: Modeling with rational functions- Analyzing structure word problem: pet store (1 of 2)
- Analyzing structure word problem: pet store (2 of 2)
- Combining mixtures example
- Rational equations word problem: combined rates
- Rational equations word problem: combined rates (example 2)
- Mixtures and combined rates word problems
- Rational equations word problem: eliminating solutions
- Reasoning about unknown variables
- Reasoning about unknown variables: divisibility
- Structure in rational expression
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Analyzing structure word problem: pet store (2 of 2)
Sal solves a word problem about the unknown number of bears, cats, and dogs in a pet store. This is part 2 where Sal uses an algebraic reasoning. Created by Sal Khan.
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When he works out b/c+d+b, couldn't he have plugged in numbers that fit the formula c>d>b?
I used c=3, d=2, and b=1.
b/c+d+b = 1/3+2+1 = 1/6
1/6<1/3
I have used this way before and it seems to work and be easier.
Is his way a more full-proof way or is my way fine?(21 votes)
Well, this is an algebra problem. Sal expects you to solve it using algebraic expressions. That's the point of this video. Though your way of solving is very useful (I use it too), it is not an algebraic solution. So no, your way is not fine, at least in this case.(3 votes)
When you analyse something.
So you look at something and try to work out how it works.(5 votes)
If the values are negative then this ratio is larger than one third.(2 votes)
How can the number of animals be negative?(6 votes)
i think there is a bit easier way to think about it, in general. ill show it here.
we are given that c > d > b. so, its clear that c > b and d > b
1. if c is greater than b, than we could say that c = b + k, where k is positive number
2. if d is greater than b, than we could say that d = b + j, where j is positive number
now we can substitude c and d: b/(c+b+d) as b/(b+k + b + b+j)
rearrange and sum the b's: b/(b+k + b + b+j) = b/(3b + k + j)
and now its simple, we know that b/3b > b/(3b + k + j) because adding positive numbers to denominator lowers the output.
b/3b = 1/3, so we get
1/3 > b/(3b + k + j)
substitude b and d back into denominator, and here is our solution:
1/3 > b/(c+b+d)(3 votes)
Yes, that is a valid approach as well. You used the fact that c > b and d > b to express c and d in terms of b, and then substituted those expressions in the denominator. From there, you used the fact that adding positive numbers to the denominator of a fraction lowers its value, and compared the resulting fraction to 1/3. Finally, you substituted the expressions for c and d back into the denominator to get the final result. Well done!(1 vote)
- How did Sal know that d = 1/3 of the rectangle in his visual analysis?
Couldn't d be a little less than 1/3 but greater than b?
Could anyone explain? -- NAZ(1 vote)dcan be1/3. We don't know what it is exactly. All we really know is thatdmust be less than1/2. The biggest it could be is whenbis zero. Thendcould be almost1/2andccould be a tiny bit more than1/2.(3 votes)
the last reasoning that C+b is greater than 2b, it seems very reasonable and easy, but can we express that very simple reasoning by math or sometimes we have to write the logic in words ?(1 vote)- Do you mean x+d . 2b? Either way, it kind of already is an algebraic argument using inequalities. Basically we are given c>d>b from the start, so we can use that. so we can then use c+1 > b+1 and c+2>b+2 and keep goign for any number and eventually get c+b > b+b(2 votes)
- kindly judge this
b/(b+c+d) = 1/ ( (d+c)/b) +1 )
b,c and d should be integers thus min b is 1, min d is 2 and min c is 3
hence: the denominator expression ((b+c)/d)+1 min value is 6
the whole expression max value is 1/6, means the expression in subject must be lesser than or equal 1/6 which in all cases lesser than 1/3
whatever, Sal's way is better, because it covers the case of non-integers values of a,b and c. such proposal only allow us to put min values and tackle the way like i did.(1 vote)- Your reasoning seems correct, but it is important to note that it only works for the specific values of b, c, and d that you provided. Sal's approach is more general and can be applied to any values of b, c, and d. Additionally, his approach uses algebraic manipulation, which is a more formal and rigorous method for comparing fractions with different denominators.(1 vote)
I am a little confused about one of the practice problems. A>B>C, and it asks the relationship between B+C and 2A-B. The analytical solution says B+C<2A-B, but when I try it with real sets of numbers I get different results. Take {A,B,C} to be the sets {3,2,1} and {4,3,2} they produce different equality relationships.(1 vote)- Maybe that's the problem where a>(b+c) is given in the problem.(1 vote)
i made it using infinity... much more complicated than Sal but here it goes.
"b tends to infinity,
as d>b, d tends to infinity+1,
and as c>d>b, c tends to infinity+2
so the denominator will have a bigger infinity than 3x(infinity)
thus b/(b+c+d) will always be <1/3."
is this reasoning correct? am i commiting conceptual mistakes?
thanks in advance(1 vote)- Your reasoning is correct in that the denominator, which tends to infinity + 3 as b tends to infinity, is much larger than 3 times infinity, so the fraction b/(b+c+d) tends to 0 as b tends to infinity. Therefore, b/(b+c+d) will always be less than 1/3.
However, it's important to note that this reasoning relies on the assumption that c and d tend to infinity at a faster rate than b. While this may be true in this particular problem, it's not necessarily always true in other situations. So, it's important to be careful when using infinity in mathematical reasoning.(1 vote)
Here is my way of solving this problem:
Takeb/(c+d+b)divided byband we have1/[(c/b)+(d/b)+1]. Becausec>bthenc/b > 1.d>bthend/b > 1too. 2 numbers bigger than 1 plus 1 will bigger than 3. And we have two rational numbers with the same numerator1. The one have bigger denominator is smaller.1/[(c/b)+(d/b)+1]<1/3=>b/(c+d+b).(1 vote)- Your approach is correct and similar to the approach taken by Sal in the original solution. By dividing both the numerator and denominator of the original expression by b, you obtained a new expression of the form 1/[(c/b)+(d/b)+1].
Then, by observing that (c/b) and (d/b) are both greater than 1, you correctly noted that their sum plus 1 is greater than 3. Therefore, 1/[(c/b)+(d/b)+1] is less than 1/3.
Finally, you concluded that b/(c+d+b) is greater than 1/3, which is the same as saying that it is less than or equal to 1/3 (since it is a positive fraction). Well done!(1 vote)
Video transcript
In the last video, we made
a visual argument as to why this expression has
to be less than 1/3, and this expression
we already figured out is the fraction that are bears. Now we will make an
algebraic argument, or I could call it
an analytic argument. And to make this
argument, I'm going to leave this expression-- we
know this is the fraction that are bears-- and I'm going to
write this 1/3 in a form that looks a lot like
this, and then based on the information we have,
we can directly compare them. So how can I write 1/3? Maybe with the b as a numerator. Well, 1/3 is the same
thing as b over 3b, which is the exact same thing
as b over b plus b plus b. So now, this is
looking pretty similar. The only difference between this
expression right over here, b over c plus d plus
b and b over b plus b plus b is that our
denominators are different. And the only difference in our
denominators, this denominator has a c plus d here, while
this has a b plus b over here. Now, we have to ask
ourselves a question. What is larger? Is c plus d larger
than b plus b? And I encourage you to
pause that and think about that for a second. Well, yes. We already see right over here. It was given to us
that c is greater than d that is greater
than b, so both c and d are greater than b. So c plus d is definitely going
to be greater than b plus b. So this denominator right
over here is greater, so this has a
larger denominator. This right over here has
a smaller denominator. And since we know this
has a larger denominator, this has a smaller
denominator, they have the exact same
numerator-- they both have b as a numerator-- we know
that this whole thing must be a smaller quantity. If you have the same numerator
but one expression has a larger denominator, it must be smaller. Wait, so how does that work? Well, just remember. I mean, just imagine. You have the same numerator,
what's going to be bigger, a over 7 or a over 5? Well here, you're
dividing a by 7. You're dividing into many
more chunks than over here, so this right over
here is smaller. This right over here is larger. So this is the larger. This right over here is smaller. So the same numerator, the
larger the denominator, the smaller the
quantity is going to be. So going back to the
original question, this is the smaller quantity,
and this right over here, 1/3, is the larger quantity.