Quadratic Functions Word Problem, It's Tricky

[Nienstien]

Homework Statement

Ok, the problem is you have a 30cm wire, you cut it in two and you make one into a square. The second part is a rectangle with a 2:1 side ratio. I need to find the perimeter of each of the two shapes if the outcome sum of areas were to be a minimum.

The Attempt at a Solution

I did all the algebra and such spent about 2 hours logically trying to crack this problem then I figured I had to make some sort of quadratic function.
I just about thought I was finished when I had these two functions:

A(square)=L(square)^2
A(rectangle)=1/2L(rectangle)^2

So from this I figured in order to have a minimum area i must put ALL of the wire into the rectangle however that is not the case. My teacher says that this must be proven.

I think the answer has to do with:

30=L(wire)=P(square)+P(rectangle)=4L(Square)=3L(Rectangle)

Any suggestions?

I don't necessarily want the answer as much as I would like to understand this kind of problem. I'm in grade 10, thanks.

 

[Tedjn]

Small correction:

30=L(wire)=P(square)+P(rectangle)=4L(Square)+3L(Rectangle)

From this, you can get rid of one pesky variable. For example, you can solve for L(Square) in terms of L(Rectangle) or the other way around. Then, we can substitute into what we are trying to minimize, which is total area:

A(Square) + A(Rectangle)

This will be in terms of one variable we can vary, namely L(Square) or L(Rectangle) depending on which variable you eliminated. Get to this point, and we can continue.

 

[Nienstien]

Ok, at this point i solved for L:

L(rectangle)=10-4/3L(square)

Now I have converted A(rectangle) to be in terms of L(square)
Since we have basically eliminated L(rectangle) I'm just going to use "L" which means L(Square)

$$A_{Rectangle}=\frac{800}{9}L^{2}-\frac{40}{3}L+50$$

Edit: Took the A(rectangle) and A(square) functions and added them together I don't know if that's how it works but I got:

$$A_{Total}=\frac{809}{9}(L-\frac{60}{809})^{2}+\frac{40050}{809}$$

Don't know what to do from here.
Edit2: Cleaned it up a bit, simplified fractions.

 

[zgozvrm]

What level of math are you studying?

 

[zgozvrm]

Do you know how to find the derivative of a function?

 

[Nienstien]

I'm in grade 10 supposedly studying grade 10 math. I'm in enriched though so my teacher likes to throw stuff she never though us on our assignments. I know a little about derivatives and limits but that's all self-learning and I wouldn't much trust it >.<

 

[zgozvrm]

Are you allowed to use a graphing calculator, programming, or Excel?

Or, is she asking you to solve this "by hand?"

 

[zgozvrm]

The problem is very simple when using calculus...

 

[Mentallic]

You don't need calculus, it's a quadratic.

For the general quadratic $$y=ax^2+bx+c$$, the maximum/minimum is located at $$x=\frac{-b}{2a}$$.

Since you found the total area in terms of L and you were asked to find for what value of L the area is a minimum, then use the above approach to find for what L the area is minimum, then the area is found by substituting that value of L into the total area formula.

 

[zgozvrm]

You don't need calculus, it's a quadratic.

For the general quadratic $$y=ax^2+bx+c$$, the maximum/minimum is located at $$x=\frac{-b}{2a}$$.

Since you found the total area in terms of L and you were asked to find for what value of L the area is a minimum, then use the above approach to find for what L the area is minimum, then the area is found by substituting that value of L into the total area formula.

True, but to some, calculus makes more sense. Differentiating [itex]ax^2+bx+c[/tex] results in [itex]2ax+b[/tex].
Set that equal to zero, and solve for x. You get the same thing:
$$x=\frac{-b}{2a}$$

 

[Mentallic]

Well it obviously can't make more sense to a student that doesn't even know calculus yet.

Yes you get the same result either from taking its derivative and setting it equal to 0 or looking at the quadratic formula and realizing that since the parabola is symmetrical about it's axis, and the roots are $$x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$$ then the centre is at $$x=\frac{-b}{2a}$$

 

[zgozvrm]

Well it obviously can't make more sense to a student that doesn't even know calculus yet.

Agreed ... that's why I asked.

 

[Mentallic]

Do you know how to find the derivative of a function?

I know a little about derivatives and limits but that's all self-learning and I wouldn't much trust it >.<

The problem is very simple when using calculus...

 

[HallsofIvy]

Well, I know Calculus, and I still prefer to find max or min for a quadratic by completing the square!

 

[zgozvrm]

I use completing the square to find factors of a quadratic.
How does it give you max or min?

 

[Nienstien]

Ok, at this point i solved for L:

L(rectangle)=10-4/3L(square)

Now I have converted A(rectangle) to be in terms of L(square)
Since we have basically eliminated L(rectangle) I'm just going to use "L" which means L(Square)

$$A_{Rectangle}=\frac{800}{9}L^{2}-\frac{40}{3}L+50$$

Edit: Took the A(rectangle) and A(square) functions and added them together I don't know if that's how it works but I got:

$$A_{Total}=\frac{809}{9}(L-\frac{60}{809})^{2}+\frac{40050}{809}$$

Don't know what to do from here.
Edit2: Cleaned it up a bit, simplified fractions.

I am expected to solve by completing the square but I do know enough calculus to get by with this type of question. I want to know if my answer sounds about right. I ended up taking

$$A_{Square}=L^{2}$$

and

$$A_{Rectangle}=LW$$

I solved for W algebraically then substituted it in.
$$W=15-\frac{2}{3}L$$
Result:

$$A_{Rectangle}=\frac{800}{9}L^{2}-\frac{40}{3}L+50$$

Then I added the two together:

$$A_{Rectangle}=\frac{809}{9}L^{2}-\frac{40}{3}L+50$$

And I got:

$$A_{Total}=\frac{809}{9}(L-\frac{60}{809})^{2}+\frac{40050}{809}$$

I think somewhere along these lines I screwed it up since if L=60/809 then using my W equation W=14.95 which is way to much considering the entire wire is 30 cm.

 

[zgozvrm]

Consider this:

Let one side of the square be represented by S.
Let the one short sides of the rectangle be represented by R.

How do you find the area of the square?
How do you find the area of the rectangle?

How do you find the perimeters of the square and the rectangle?

 

[Mentallic]

I use completing the square to find factors of a quadratic.
How does it give you max or min?

Since you must complete the square to give the quadratic formula, they're pretty much the same thing.
Like I said, when given the quadratic formula $$x=\frac{-b\pm\sqrt{\Delta}}{2a}$$ then you can find the turning point or max/min of the parabola by considering that its roots lie on either end of it, so $$x=\frac{-b}{2a}$$.

Equivalently, we can complete the square on the general quadratic to end up with $$(x+\frac{b}{2a})^2=\frac{\Delta}{4a^2}$$

And so we have the parabolic function $$y=(x+\frac{b}{2a})^2-\frac{\Delta}{4a^2}$$
Now, we remember that for any real number x, $$x^2\geq 0$$ so the minimum that $$(x+\frac{b}{2a})^2$$ can be is 0, and this occurs when $$x=\frac{-b}{2a}$$. So when x is this value, $$y=\frac{-\Delta}{4a^2}$$ and this would be the minimum/maximum point.

Im expected to solve by completing the square but I do know enough calculus to get by with this type of question. I want to know if my answer sounds about right. I ended up taking

$$A_{Square}=L^{2}$$

and

$$A_{Rectangle}=LW$$

I solved for W algebraically then substituted it in.
$$W=15-\frac{2}{3}L$$
Result:

$$A_{Rectangle}=\frac{800}{9}L^{2}-\frac{40}{3}L+50$$

Then I added the two together:

$$A_{Rectangle}=\frac{809}{9}L^{2}-\frac{40}{3}L+50$$

And I got:

$$A_{Total}=\frac{809}{9}(L-\frac{60}{809})^{2}+\frac{40050}{809}$$

I think somewhere along these lines I screwed it up since if L=60/809 then using my W equation W=14.95 which is way to much considering the entire wire is 30 cm.

You should be getting $$W=5-\frac{2}{3}L$$

If you let the short side of the rectangle be W, then the perimeter of the whole rectangle is 2(W)+2(2W)=6W. The area is (2W)(W)=2W² and we have that 4L+6W=30 since this is the perimeter of both the square and rectangle consists of the 30cm of wiring.

Solving for W gives $$W=5-\frac{2}{3}L$$

 

[zgozvrm]

Yeah, I still don't see why you'd need to complete the square.

Starting with (from the perimeters)
$$W = 5 - \frac{2}{3} L$$

and the sum of the areas of the square and rectangle
$$L^2 + 2W^2$$


Substituting for W, you get
$$L^2 + 2 \left( 5 - \frac{2}{3} L \right) ^2$$

Simplifying, you get

$$L^2 + 2 \left( 25 - \frac{20}{3} L + \frac{4}{9} L^2 \right)$$

$$= L^2 +50 - \frac{40}{3} L + \frac{8}{9} L^2$$

$$= \frac{17}{9} L^2 - \frac{40}{3} L + 50 = \frac{1}{9} (17L^2 -120L +450)$$


There's your quadratic ... no completing the square necessary!

 

[Mentallic]

Did you read the entire question? It asks what the side lengths are of the square and rectangle when their areas are a minimum. How are you going to find the minimum of that quadratic without calculus?

 

[zgozvrm]

How are you going to find the minimum of that quadratic without calculus?

According to your first post (post #9):

You don't need calculus, it's a quadratic.

For the general quadratic $$y=ax^2+bx+c$$, the maximum/minimum is located at $$x=\frac{-b}{2a}$$.

So,
$$x = \frac{40/3}{2 \times 17/9}=\frac{40/3}{34/9} = \frac{40}{3} \times \frac{9}{34} = \frac{60}{17}$$

 

[Mentallic]

And since "completing the square" was thrown out there, I clarified in post #18 that manipulating the quadratic formula or completing the square are essentially doing the same thing.

Right, so the minimum total area occurs when the side length of the square, L=60/17.

 

[zgozvrm]

I still don't see the need for completing the square.

If you already have the quadratic, you can find the min. value from that, so why find the factors by completing the square?

Why do the extra work?

 

[Mentallic]

Please read my reply to you in post #18.
Completing the square has benefits other than just finding the factors of the quadratic.

 

[zgozvrm]

Please read my reply to you in post #18.
Completing the square has benefits other than just finding the factors of the quadratic.

Okay, I've read it and now I've re-read it. You're still coming up with the same answer, but you have to do more work to get there.

What are some of the other "benefits" you mentioned?

 

[Mentallic]

Okay, I've read it and now I've re-read it. You're still coming up with the same answer, but you have to do more work to get there.

More work? The only extra work is the necessary work of completing the square to answer the question. You haven't answered the question:

There's your quadratic ... no completing the square necessary!

But then to do so you needed to use my idea which is derived from completing the square:

According to your first post (post #9):

What are some of the other "benefits" you mentioned?

It's in my post which you've re-read a few times now:

And so we have the parabolic function $$y=(x+\frac{b}{2a})^2-\frac{\Delta}{4a^2}$$
Now, we remember that for any real number x, $$x^2\geq 0$$ so the minimum that $$(x+\frac{b}{2a})^2$$ can be is 0, and this occurs when $$x=\frac{-b}{2a}$$. So when x is this value, $$y=\frac{-\Delta}{4a^2}$$ and this would be the minimum/maximum point.

If it's still not clear on what is happening, then I implore you to answer the original question in its entirety, provided you don't use calculus. If you find the answer then you would've most likely used the quadratic formula or completed the square to do so.

 

[zgozvrm]

More work? The only extra work is the necessary work of completing the square to answer the question.

That's still more work. Any additional work is MORE work, no matter how small.

But then to do so you needed to use my idea which is derived from completing the square

By "your idea," I assume you mean finding the minimum value of a quadratic by using
$$x = \frac{-b}{2a}$$

correct?

Although it was derived by completing the square (it can also be derived with calculus by differentiation), you don't need to prove it to use it - if we had to "invent the wheel" each time we use a mathematical property, theorem, law, etc., we'd get nowhere, fast. That would be similar to having to prove the quadratic formula, or the Pythagorean Theorem every time we use one of those formulas.

It's sufficient to say that the x-value for the minimum value of a quadratic function can be found at
$$x = \frac{-b}{2a}$$

Then plug that value back into the quadratic to get the actual min. value.


So, the question remains, why go through the extra work of finding the factors of a quadratic, when all you have to do is use the values that are staring you right in the face?

 

[eumyang]

I still don't see the need for completing the square.

If you already have the quadratic, you can find the min. value from that, so why find the factors by completing the square?

Why do the extra work?

Completing the square is not just used for finding the zeros of a quadratic equation. It can also be used to transform a quadratic from standard form:
y = ax² + bx + c
into vertex form:
y = a(x - h)² + k.
I think this is what Mentallic was getting at.

Here's an example:
y = -4x² + 16x - 23
y = -4(x² - 4x) - 23
y = -4(x² - 4x + 4 - 4) - 23 [here is where we complete the square]
y = -4(x² - 4x + 4) + 16 - 23
y = -4(x - 2)² - 7

Once you have the quadratic in vertex form, you can also quickly identify your max/min point. Of course, there is the second way, where you find h by evaluating
h = -b/2a
and k by evaluating
k = c - ah² or k = f(h).

A math teacher may require that students be able to find the vertex using either method. The OP did say that he's to solve his problem by completing the square:

Im expected to solve by completing the square but I do know enough calculus to get by with this type of question.

Hope this helps.

 

[Mentallic]

How did you get $$x=\frac{-b}{2a}$$? Using calculus? Sorry, this isn't allowed. Whatever you do to find the answer needs to be repeatable by the OP which hasn't learned calculus yet.
At first you didn't know any other way besides using calculus to find the turning point of a quadratic. I had to show it to you. Now let's assume that if it wasn't general knowledge for you to know that, then it isn't general knowledge for the OP either. How is the OP going to solve this problem then? By completing the square.

And like I've already said many times, we aren't finding the factors of the quadratic. When we complete the square, even though a little more work will lead us to the factors of the quadratic, we don't need to go there, we are just using the fact that for a quadratic which has been transformed into a complete square, then

$$y=a(x-c)^2+k$$ has a turning point at x=c and the value at this point is k. That is all. We aren't going to solve for x after plugging in y=0.

 

[Mentallic]

Yes, thankyou eumyang. I hope that's where the confusion was so we can finally settle this

 

[eumyang]

Yes, thankyou eumyang. I hope that's where the confusion was so we can finally settle this

You're welcome!

@zgozvrm: there is yet another way to find the value -b/2a. You can take the two forms of the quadratic (standard form and vertex form) and set them equal to each other:

ax² + bx + c = a(x - h)² + k

Expand the right hand side, and then:
* set the coefficients of the x terms equal to each other, and
* set the constant terms equal to each other,
to solve for h and k. You should get
h = -b/2a and k = c - ah². (Now note, this is another way to derive the formulas for h and k. We wouldn't be repeating this every time we want to solve a specific problem regarding quadratics.)

 

[zgozvrm]

Completing the square is not just used for finding the zeros of a quadratic equation. It can also be used to transform a quadratic from standard form:
y = ax² + bx + c
into vertex form:
y = a(x - h)² + k.

Once you have the quadratic in vertex form, you can also quickly identify your max/min point.

I get that, but my point is this: once shown that h = -b/2a, there is no longer any need to complete the square for additional problems.

In your example of -4x² + 16x - 23, we can clearly see that h = -16 / 2(-4) = 2 so why complete the square to get y = -4(x - 2)² - 7, which shows the same thing? ... it's just extra work.



Besides, in post #9, you stated,

For the general quadratic $$y=ax^2+bx+c$$, the maximum/minimum is located at $$x=\frac{-b}{2a}$$

without mentioning how you got there, until HallsofIvy mentioned "completing the square" and I asked about that method. So, the assumption here, is that you can use this identity without completing the square for quadratics.


Also, the OP didn't state that he was to solve by completing the square until after the topic was brought up.

 

[Mentallic]

... it's just extra work.

The OP said he's already spent 2 hours on this problem. Saving time should be the last of his worries in my opinion.

without mentioning how you got there

I didn't think I would need to because he is studying quadratics and would have stumbled across this result many times by now. On the other hand, you didn't know it so I acted accordingly by showing you how it works.

I use completing the square to find factors of a quadratic.
How does it give you max or min?

Then later down the track the OP said it needs to be solved by completing the square, and you're saying it's just extra work to do so since the result has already been given here.

Obviously the extra work isn't going to go to waste.

 

[Tedjn]

I don't want to get into a huge argument but just wanted to throw in my two cents. The OP's original question is essentially solved already anyway.

The fact that x = -b/2a is the min or max of a quadratic seems to me like one of those useful facts that you pick up when working on math competitions. It's always a good idea to know that it does indeed come out of completing the square (or calculus). And, I do think that is a nontrivial fact in that many people I know (granted, not mathematicians) wouldn't know this off the top of their head.

More important is what is gained from going through the derivations. The reason why completing the square works well in giving a min/max is because you can work the equation into the form (something)^2 + something, and we can use the fact that (something^2) >= 0. In fact, this is in many ways the most basic and important inequality, the starting point of so many other inequalities, so it's nice to see it used.